Chapter 7 Permutations And Combination

Given:

Number of boys in the class = $27$

Number of girls in the class = $14$

To Find:

The number of ways to select 1 boy and 1 girl from the class.

Solution:

The selection requires choosing 1 boy from the available boys and 1 girl from the available girls.

The number of ways to select 1 boy from $27$ boys is the number of combinations of $27$ items taken 1 at a time, which is given by $\binom{27}{1}$.

The number of ways to select 1 girl from $14$ girls is the number of combinations of $14$ items taken 1 at a time, which is given by $\binom{14}{1}$.

Since the selection of a boy and the selection of a girl are independent events, the total number of ways to select 1 boy and 1 girl is the product of the number of ways to perform each selection. This is based on the Fundamental Principle of Counting.

Thus, the teacher can make the selection in $378$ ways.

The numbers between $99$ and $1000$ are the three-digit numbers, ranging from $100$ to $999$. There are a total of $999 - 100 + 1 = 900$ three-digit numbers.

(i) Numbers having 7 in the units place:

We are looking for three-digit numbers where the units digit is $7$. A three-digit number can be represented as H T U, where H is the hundreds digit, T is the tens digit, and U is the units digit.

For a three-digit number:

The hundreds digit (H) can be any digit from $1$ to $9$ (since the number must be at least $100$). There are $9$ choices for H.

The tens digit (T) can be any digit from $0$ to $9$. There are $10$ choices for T.

The units digit (U) must be $7$. There is $1$ choice for U.

By the Fundamental Principle of Counting, the total number of such numbers is the product of the number of choices for each digit.

So, there are $90$ numbers between $99$ and $1000$ having $7$ in the units place.

(ii) Numbers having at least one of their digits 7:

It is easier to find the total number of three-digit numbers and subtract the number of three-digit numbers that have no digit $7$.

Total number of three-digit numbers (from $100$ to $999$):

Hundreds digit (H): $1, 2, ..., 9$ (9 choices)

Tens digit (T): $0, 1, ..., 9$ (10 choices)

Units digit (U): $0, 1, ..., 9$ (10 choices)

Number of three-digit numbers with no digit $7$:

Hundreds digit (H): Can be any digit from $1$ to $9$ except $7$. The choices are $1, 2, 3, 4, 5, 6, 8, 9$. There are $8$ choices for H.

Tens digit (T): Can be any digit from $0$ to $9$ except $7$. The choices are $0, 1, 2, 3, 4, 5, 6, 8, 9$. There are $9$ choices for T.

Units digit (U): Can be any digit from $0$ to $9$ except $7$. The choices are $0, 1, 2, 3, 4, 5, 6, 8, 9$. There are $9$ choices for U.

The number of three-digit numbers having at least one digit $7$ is the total number of three-digit numbers minus the number of three-digit numbers with no digit $7$.

So, there are $252$ numbers between $99$ and $1000$ having at least one of their digits $7$.

Given:

The diagram consists of a central region and three smaller triangles surrounding it.

Available colours: Red, Blue, Green (Total 3 colours).

Condition: No two adjacent regions have the same colour.

To Find:

The number of ways the diagram can be coloured satisfying the given conditions.

Solution:

Let's identify the regions. We have a central region and three surrounding triangular regions. The central region is adjacent to all three surrounding triangles. The three surrounding triangles are not adjacent to each other.

We can colour the regions step-by-step:

Step 1: Colour the central region.

There are $3$ available colours (Red, Blue, Green). So, the central region can be coloured in $3$ ways.

Step 2: Colour the three surrounding triangles.

Each surrounding triangle is adjacent to the central region. Therefore, its colour must be different from the colour of the central region.

Since there are $3$ colours in total, and $1$ colour is used for the central region, there are $3 - 1 = 2$ choices for the colour of each surrounding triangle.

Also, the three surrounding triangles are not adjacent to each other, so the colour choice for one surrounding triangle does not affect the colour choice for another surrounding triangle, other than the restriction imposed by the central region's colour.

For the first surrounding triangle, there are $2$ colour choices.

For the second surrounding triangle, there are $2$ colour choices.

For the third surrounding triangle, there are $2$ colour choices.

By the Fundamental Principle of Counting, the total number of ways to colour the entire diagram is the product of the number of ways to colour each region independently.

Thus, the diagram can be coloured in $24$ ways subject to the given conditions.

Given:

Total number of children = $5$.

The children are to be arranged in a line.

We consider two particular children among them.

To Find:

(i) The number of arrangements where the two particular children are always together.

(ii) The number of arrangements where the two particular children are never together.

Solution:

Let the five children be $C_1, C_2, C_3, C_4, C_5$. Let the two particular children be $P_1$ and $P_2$. The remaining children are $R_1, R_2, R_3$.

(i) Two particular children are always together:

Treat the two particular children ($P_1$ and $P_2$) as a single unit or a block. Now we have this unit and the other $5-2=3$ children to arrange. So, we are arranging a total of $3+1=4$ entities (the block of $P_1, P_2$, and the 3 other children $R_1, R_2, R_3$).

The number of ways to arrange these $4$ entities is the permutation of $4$ items, which is $4!$.

Within the block of the two particular children ($P_1, P_2$), they can arrange themselves in $2!$ ways (either $P_1$ followed by $P_2$ or $P_2$ followed by $P_1$).

By the Fundamental Principle of Counting, the total number of arrangements where the two particular children are always together is the product of the number of ways to arrange the units and the number of ways to arrange children within the block.

So, there are $48$ ways for the children to be arranged such that the two particular children are always together.

(ii) Two particular children are never together:

The total number of ways to arrange $5$ children in a line without any restrictions is the permutation of $5$ items, which is $5!$.

The number of arrangements where the two particular children are never together is the total number of arrangements minus the number of arrangements where they are always together (calculated in part (i)).

So, there are $72$ ways for the children to be arranged such that the two particular children are never together.

Given:

The word is AGAIN.

The letters are A, G, A, I, N.

The letters arranged in alphabetical order are A, A, G, I, N.

To Find:

The 49^th word when all permutations of the letters of AGAIN are arranged in dictionary order.

Solution:

The letters of the word AGAIN are A (appears 2 times), G (1 time), I (1 time), N (1 time). There are a total of 5 letters.

We arrange the letters in alphabetical order: A, A, G, I, N.

We list the permutations in dictionary order:

1. Words starting with A:

If the first letter is A, the remaining letters are A, G, I, N. These 4 letters can be arranged in $\frac{4!}{2!1!1!1!} = \frac{24}{2} = 12$ ways. However, the original word has A appearing twice. Let's re-evaluate carefully.

The distinct letters are A, G, I, N. Alphabetical order is A, G, I, N.

Words starting with A:

Fix the first letter as A. The remaining letters are A, G, I, N. The permutations of these 4 letters, considering the repeated A, are $\frac{4!}{1!1!1!1!} = 24$.

So, there are 24 words starting with A.

These words are ranked from 1 to 24.

2. Words starting with G:

Fix the first letter as G. The remaining letters are A, A, I, N. The permutations of these 4 letters, considering the repeated A, are $\frac{4!}{2!1!1!} = \frac{24}{2} = 12$.

So, there are 12 words starting with G.

These words are ranked from $24 + 1 = 25$ to $24 + 12 = 36$.

3. Words starting with I:

Fix the first letter as I. The remaining letters are A, A, G, N. The permutations of these 4 letters, considering the repeated A, are $\frac{4!}{2!1!1!} = \frac{24}{2} = 12$.

So, there are 12 words starting with I.

These words are ranked from $36 + 1 = 37$ to $36 + 12 = 48$.

We need to find the 49^th word. The first 48 words start with A, G, or I.

The 49^th word must be the first word starting with the next letter in alphabetical order, which is N.

4. Words starting with N:

Fix the first letter as N. The remaining letters are A, A, G, I. To find the first word starting with N in dictionary order, we arrange the remaining letters (A, A, G, I) in alphabetical order: A, A, G, I.

So, the first word starting with N is N followed by the alphabetically arranged remaining letters.

This is the word that comes immediately after all words starting with A, G, and I in the dictionary order.

The rank of this word is $48 + 1 = 49$.

Therefore, the 49^th word is NAAGI.

Given:

Number of Mathematics books = $3$

Number of History books = $4$}

Number of Chemistry books = $3$

Number of Biology books = $2$

Condition: All books of the same subjects must be arranged together.

To Find:

The total number of ways to arrange the books on a shelf under the given condition.

Solution:

Since all books of the same subject must be together, we can treat each group of books of the same subject as a single unit.

We have the following units:

1. Unit of 3 Mathematics books

2. Unit of 4 History books

3. Unit of 3 Chemistry books

4. Unit of 2 Biology books

We need to arrange these $4$ distinct units on the shelf. The number of ways to arrange $4$ distinct units is given by the permutation of $4$ items, which is $4!$.

Now, consider the arrangements of books within each unit:

The 3 Mathematics books can be arranged among themselves in $3!$ ways.

The 4 History books can be arranged among themselves in $4!$ ways.

The 3 Chemistry books can be arranged among themselves in $3!$ ways.

The 2 Biology books can be arranged among themselves in $2!$ ways.

By the Fundamental Principle of Counting, the total number of ways to arrange the books such that all books of the same subject are together is the product of the number of ways to arrange the subject units and the number of ways to arrange the books within each unit.

So, there are $6912$ ways to arrange the books on the shelf such that all books of the same subjects are together.

Given:

Total questions in Part A = $6$

Total questions in Part B = $7$

Total questions to be answered = $10$

Condition: At least $4$ questions must be chosen from each of Part A and Part B.

To Find:

The number of ways the student can choose $10$ questions according to the given conditions.

Solution:

Let $n_A$ be the number of questions chosen from Part A and $n_B$ be the number of questions chosen from Part B.

We are given that the total number of questions chosen is $10$:

The constraints are:

Also, the number of questions chosen from each part cannot exceed the total number of questions in that part:

From $n_A + n_B = 10$, we have $n_B = 10 - n_A$. Substituting this into the inequality $n_B \ge 4$, we get:

Combining the constraints on $n_A$: $n_A \ge 4$ and $n_A \le 6$. Thus, the possible values for $n_A$ are $4, 5,$ or $6$.

We consider each case for $(n_A, n_B)$:

Case 1: $n_A = 4$

If $n_A = 4$, then $n_B = 10 - 4 = 6$. This satisfies $n_A \ge 4, n_A \le 6, n_B \ge 4, n_B \le 7$.

Number of ways to choose 4 questions from 6 in Part A = $\binom{6}{4}$.

Number of ways to choose 6 questions from 7 in Part B = $\binom{7}{6}$.

Number of ways for Case 1 = $\binom{6}{4} \times \binom{7}{6} = 15 \times 7 = 105$.

Case 2: $n_A = 5$

If $n_A = 5$, then $n_B = 10 - 5 = 5$. This satisfies $n_A \ge 4, n_A \le 6, n_B \ge 4, n_B \le 7$.

Number of ways to choose 5 questions from 6 in Part A = $\binom{6}{5}$.

Number of ways to choose 5 questions from 7 in Part B = $\binom{7}{5}$.

Number of ways for Case 2 = $\binom{6}{5} \times \binom{7}{5} = 6 \times 21 = 126$.

Case 3: $n_A = 6$

If $n_A = 6$, then $n_B = 10 - 6 = 4$. This satisfies $n_A \ge 4, n_A \le 6, n_B \ge 4, n_B \le 7$.

Number of ways to choose 6 questions from 6 in Part A = $\binom{6}{6}$.

Number of ways to choose 4 questions from 7 in Part B = $\binom{7}{4}$.

Number of ways for Case 3 = $\binom{6}{6} \times \binom{7}{4} = 1 \times 35 = 35$.

Since these cases are mutually exclusive, the total number of ways to choose the 10 questions is the sum of the ways in each case.

Thus, the student can choose the 10 questions in $266$ ways.

Given:

Number of men = $m$

Number of women = $n$

Condition: No two women sit together ($m > n$).

The men and women are to be seated in a row.

To Prove:

The number of ways in which they can be seated such that no two women sit together is $\frac{m! (m+1)!}{(m-n+1)!}$.

Proof:

To ensure that no two women sit together, we first arrange the men. There are $m$ distinct men, and they can be arranged in a row in $m!$ ways.

When the $m$ men are arranged in a row, they create $m+1$ possible spaces where the women can be seated so that no two women are adjacent. These spaces are before the first man, between any two consecutive men, and after the last man.

Let M represent a man. The arrangement of $m$ men creates spaces (\_) like this:

$\_$ M $\_$ M $\_$ M $\_ \dots \_$ M $\_$

There are $m$ men and $m+1$ spaces.

We need to place $n$ women in these $m+1$ spaces such that no two women are in the same space (since they are distinct individuals) and each space can hold at most one woman to satisfy the condition of no two women sitting together. Since $m > n$, there are more spaces than women ($m+1 \ge n+2 > n$), so it is possible to place the women in distinct spaces.

First, we choose $n$ spaces out of the $m+1$ available spaces where the women will be seated. The number of ways to choose $n$ spaces from $m+1$ spaces is given by the combination formula $\binom{m+1}{n}$.

Once the $n$ spaces are chosen, the $n$ distinct women can be arranged in these $n$ chosen spaces in $n!$ ways.

By the Fundamental Principle of Counting, the total number of ways to seat the men and women according to the given condition is the product of the number of ways to arrange the men and the number of ways to place the women in the chosen spaces.

Now, we expand the combination term $\binom{m+1}{n}$:

Substitute this back into the expression for the total number of ways:

Cancel out the $n!$ term from the numerator and the denominator:

This is the required expression.

Hence, the number of ways in which $m$ men and $n$ women can be seated in a row so that no two women sit together, given $m > n$, is $\frac{m! (m+1)!}{(m-n+1)!}$.

Given:

Number of married couples = $3$

Total number of people = $3 \text{ couples} \times 2 \text{ people/couple} = 6$

Total number of seats in the row = $6$

To Find:

1. The number of ways they can be seated if spouses are seated next to each other.

2. The number of ways they can be seated if all the ladies sit together.

Solution - Part 1: Spouses are seated next to each other

Let the three married couples be denoted as $C_1, C_2, C_3$. Since spouses must sit together, we can treat each couple as a single unit.

We have $3$ such units (Couple 1, Couple 2, Couple 3) to arrange in a row.

The number of ways to arrange these $3$ distinct units is the permutation of $3$ items, which is $3!$.

Within each couple unit $(M, W)$, the two individuals can arrange themselves in two ways: $M$ followed by $W$ or $W$ followed by $M$. The number of ways to arrange the two individuals within a couple is $2!$.

Since there are $3$ couples, and the arrangements within each couple are independent, we multiply the number of ways for each couple's internal arrangement.

By the Fundamental Principle of Counting, the total number of ways to seat the couples such that spouses are next to each other is the product of the number of ways to arrange the couple units and the number of ways to arrange individuals within each couple unit.

So, there are $48$ ways to seat the couples if spouses are to be seated next to each other.

Solution - Part 2: All the ladies sit together

There are $3$ ladies and $3$ men. If all the ladies must sit together, we can treat the group of $3$ ladies as a single unit or a block.

Now, we have this unit of $3$ ladies and the $3$ men to arrange in a row. This gives us a total of $3 \text{ men} + 1 \text{ unit (ladies)} = 4$ entities to arrange.

The number of ways to arrange these $4$ distinct entities (the block of ladies and the 3 men) is the permutation of $4$ items, which is $4!$.

Within the block of $3$ ladies, the ladies can arrange themselves in $3!$ ways.

By the Fundamental Principle of Counting, the total number of ways to seat the people such that all ladies sit together is the product of the number of ways to arrange the entities and the number of ways to arrange the ladies within their block.

So, there are $144$ ways to seat the people if all the ladies sit together.

Given:

Total number of families in the village = $87$

Number of families with at most 2 children = $52$ (Let this be Group A)

Number of families with more than 2 children = Total families - Families with at most 2 children = $87 - 52 = 35$ (Let this be Group B)

Number of families to be chosen for assistance = $20$

Condition: At least 18 families chosen must have at most 2 children (from Group A).

To Find:

The number of ways to choose 20 families such that at least 18 are from Group A.

Solution:

The condition "at least 18 families must have at most 2 children" means that the number of families chosen from Group A can be 18, 19, or 20.

Let $n_A$ be the number of families chosen from Group A and $n_B$ be the number of families chosen from Group B.

We must choose a total of $20$ families, so $n_A + n_B = 20$.

The condition is $n_A \ge 18$.

Since the total number of families in Group A is 52, $n_A$ can be at most 20 (as we are choosing 20 families in total). So, $18 \le n_A \le 20$.

We consider the possible cases for the number of families chosen from Group A and the corresponding number of families chosen from Group B.

Case 1: $n_A = 18$

If 18 families are chosen from Group A, then the number of families chosen from Group B must be $n_B = 20 - n_A = 20 - 18 = 2$.

This case involves choosing 18 families from 52 (Group A) and 2 families from 35 (Group B).

Number of ways for Case 1 = (Ways to choose 18 from Group A) $\times$ (Ways to choose 2 from Group B)

Case 2: $n_A = 19$

If 19 families are chosen from Group A, then the number of families chosen from Group B must be $n_B = 20 - n_A = 20 - 19 = 1$.

This case involves choosing 19 families from 52 (Group A) and 1 family from 35 (Group B).

Number of ways for Case 2 = (Ways to choose 19 from Group A) $\times$ (Ways to choose 1 from Group B)

Case 3: $n_A = 20$

If 20 families are chosen from Group A, then the number of families chosen from Group B must be $n_B = 20 - n_A = 20 - 20 = 0$.

This case involves choosing 20 families from 52 (Group A) and 0 families from 35 (Group B).

Number of ways for Case 3 = (Ways to choose 20 from Group A) $\times$ (Ways to choose 0 from Group B)

Since these three cases are mutually exclusive, the total number of ways to choose the 20 families is the sum of the number of ways in each case.

Thus, the total number of ways the choice can be made is $\mathbf{595 \times \binom{52}{18} + 35 \times \binom{52}{19} + \binom{52}{20}}$.

Given:

Total number of books of interest = $8$

Number of library tickets = $3$ (meaning he can borrow 3 books)

Two specific books: Mathematics Part I (M1) and Mathematics Part II (M2), both included in the 8 books.

Condition: M2 is not borrowed unless M1 is also borrowed.

To Find:

The number of ways the student can choose 3 books satisfying the given condition.

Solution:

Let the 8 books consist of the two specific books M1 and M2, and the remaining $8 - 2 = 6$ other books.

The condition states that M2 cannot be borrowed unless M1 is also borrowed. This means that if M2 is selected, M1 must also be selected. This condition can be analyzed by considering the possibilities involving M1 and M2 when choosing the 3 books:

There are two mutually exclusive cases that satisfy the condition:

Case 1: M2 is selected.

According to the condition, if M2 is selected, then M1 must also be selected.

So, in this case, the student selects both M1 and M2. This accounts for 2 of the 3 books to be borrowed.

The number of remaining books to be chosen is $3 - 2 = 1$.

These remaining 1 book must be chosen from the other 6 books (neither M1 nor M2).

The number of ways to choose 1 book from the 6 other books is given by the combination $\binom{6}{1}$.

Case 2: M2 is NOT selected.

If M2 is not selected, the condition is automatically satisfied (since M2 is not borrowed without M1). In this case, the student can choose the 3 books from the set of 8 books excluding M2.

The number of available books to choose from (excluding M2) is $8 - 1 = 7$. This set of 7 books includes M1 and the 6 other books.

The student needs to choose 3 books from these 7 books.

The number of ways to choose 3 books from these 7 books is given by the combination $\binom{7}{3}$.

Since Case 1 and Case 2 are mutually exclusive and cover all valid scenarios, the total number of ways to choose the 3 books is the sum of the ways in Case 1 and Case 2.

Thus, the student can choose the three books to be borrowed in $41$ ways.

Given:

Total number of different things = $n$

Number of things taken at a time for permutation = $r$}

Condition: Two specific things out of the $n$ things must occur together in the permutation.

We assume $r \ge 2$ since two specific things must occur together.

To Find:

The number of permutations of $n$ different things taken $r$ at a time such that two specific things occur together.

Solution:

Let the $n$ different things be $T_1, T_2, \dots, T_n$. Let the two specific things that must occur together be $S_1$ and $S_2$. These $S_1$ and $S_2$ are among the $n$ things.

To ensure that $S_1$ and $S_2$ always occur together in a permutation of $r$ things, we can treat them as a single unit or a block.

Step 1: Consider the unit formed by the two specific things ($S_1, S_2$).

Within this unit, the two things can arrange themselves in $2!$ ways (either $S_1$ followed by $S_2$, or $S_2$ followed by $S_1$).

Step 2: Select the remaining items for the permutation.

We need to form a permutation of $r$ things. One unit consists of the two specific things. This means we need to select $r - 2$ additional things from the remaining $n - 2$ things (which are not $S_1$ or $S_2$).

The number of ways to choose $r-2$ things from $n-2$ different things is given by the combination formula $\binom{n-2}{r-2}$.

Step 3: Arrange the selected items.

Now we have a total of $r-1$ entities to arrange: the single unit containing $S_1$ and $S_2$, and the $r-2$ individual things chosen in Step 2. These $r-1$ distinct entities can be arranged in a row in $(r-1)!$ ways.

By the Fundamental Principle of Counting, the total number of permutations satisfying the condition is the product of the number of ways for each step:

Substitute the formula for combination:

So, the total number of permutations is:

Rewrite $(r-1)!$ as $(r-1) \times (r-2)!$ and $2!$ as $2$:

Cancel out the $(r-2)!$ term from the numerator and denominator:

This can also be written using permutation notation $P(k, m) = \frac{k!}{(k-m)!}$:

So, the total number of permutations is:

Thus, the number of permutations of $n$ different things taken $r$ at a time such that two specific things occur together is $2(r-1) \frac{(n-2)!}{(n-r)!}$ or $2(r-1) P(n-2, r-2)$.

Given:

Number of bus routes between A and B = $4$

Number of bus routes between B and C = $3$

The man travels round-trip from A to C via B, which means the path is A $\to$ B $\to$ C $\to$ B $\to$ A.

Condition: He does not want to use a bus route more than once for the entire round trip.

To Find:

The number of ways the man can make the round trip under the given condition.

Solution:

A round trip from A to C via B involves four legs: A to B (outward), B to C (outward), C to B (return), and B to A (return).

Let the route taken from A to B be $R_{AB}$.

Let the route taken from B to C be $R_{BC}$.

Let the route taken from C to B be $R_{CB}$.

Let the route taken from B to A be $R_{BA}$.

The condition is that the four routes $R_{AB}, R_{BC}, R_{CB}, R_{BA}$ must all be distinct.

Step 1: Journey from A to C via B (Outward trip)

Number of ways to travel from A to B = $4$ (any of the 4 routes)

Number of ways to travel from B to C = $3$ (any of the 3 routes)

The total number of ways for the outward journey (A $\to$ B $\to$ C) is the product of the number of choices for each leg, by the Fundamental Principle of Counting.

Let the chosen route from A to B be a specific route, say $r_{AB}$, and the chosen route from B to C be a specific route, say $r_{BC}$.

Step 2: Journey from C to A via B (Return trip)

The man travels from C to B. There are 3 routes between B and C. Since he cannot use the route $r_{BC}$ again (as per the condition), he must choose one of the remaining $3 - 1 = 2$ routes for the C to B leg ($R_{CB}$).

Next, the man travels from B to A. There are 4 routes between A and B. Since he cannot use the route $r_{AB}$ again, he must choose one of the remaining $4 - 1 = 3$ routes for the B to A leg ($R_{BA}$).

The total number of ways for the return journey (C $\to$ B $\to$ A) for a given outward journey is the product of the number of choices for each leg.

Step 3: Total number of ways for the round trip

The total number of ways for the round trip is the product of the number of ways for the outward journey and the number of ways for the return journey (given the outward routes), by the Fundamental Principle of Counting.

The number of ways he can make the round trip without using a bus route more than once is 72.

The correct option is (A) 72.

Given:

Total number of men available = $7$

Total number of women available = $5$

Committee composition required: $3$ men and $2$ women.

To Find:

The number of ways to choose a committee consisting of 3 men and 2 women.

Solution:

The problem requires selecting a group of individuals for a committee, where the order of selection does not matter. This is a combination problem.

The selection of men and the selection of women are independent events.

Number of ways to choose 3 men from 7 men:

This is given by the combination formula $\binom{7}{3}$.

Number of ways to choose 2 women from 5 women:

This is given by the combination formula $\binom{5}{2}$.

By the Fundamental Principle of Counting, the total number of ways to form the committee is the product of the number of ways to choose the men and the number of ways to choose the women.

The number of ways to choose a committee consisting of 3 men and 2 women from 7 men and 5 women is 350.

The correct option is (B) 350.

Given:

The word is EAMCOT.

The letters in the word are E, A, M, C, O, T.

Total number of letters = $6$. All letters are distinct.

Identify the vowels and consonants in the word:

Vowels: E, A, O (3 vowels)

Consonants: M, C, T (3 consonants)

Condition: No two vowels are adjacent to each other in the arrangement.

To Find:

The number of distinct arrangements of the letters of the word EAMCOT such that no two vowels are adjacent.

Solution:

To ensure that no two vowels are adjacent, we can first arrange the consonants and then place the vowels in the spaces created between and at the ends of the consonants.

Step 1: Arrange the consonants.

There are 3 distinct consonants (M, C, T). The number of ways to arrange these 3 consonants in a row is $3!$.

Step 2: Identify the spaces for vowels.

When the 3 consonants are arranged, they create $3 + 1 = 4$ possible spaces where the vowels can be placed. These spaces are before the first consonant, between any two consecutive consonants, and after the last consonant.

Let C denote a consonant. The arrangement looks like: $\_$ C $\_$ C $\_$ C $\_$

There are 4 available spaces.

Step 3: Place the vowels in the spaces.

We have 3 distinct vowels (E, A, O) to place in the 4 spaces such that no two vowels are in the same space (which would make them adjacent). This means we need to choose 3 of the 4 spaces and arrange the 3 distinct vowels in those chosen spaces.

The number of ways to select 3 spaces out of 4 is $\binom{4}{3}$.

The number of ways to arrange the 3 distinct vowels in the 3 chosen spaces is $3!$.

Alternatively, the number of ways to place 3 distinct vowels into 4 distinct spaces, one vowel per space, is the number of permutations of 4 items taken 3 at a time, $P(4, 3)$.

Step 4: Combine the arrangements.

The total number of arrangements where no two vowels are adjacent is the product of the number of ways to arrange the consonants and the number of ways to place the vowels in the spaces.

The number of arrangements in which no two vowels are adjacent is 144.

The correct option is (B) 144.

Given:

Total number of different letters available = $10$

Length of the words to be formed = $5$ letters

Repetition of letters is allowed.

To Find:

The number of words of five letters formed from the given ten letters such that at least one letter is repeated.

Solution:

The total number of words that can be formed using 10 different letters, taking 5 at a time, with repetition allowed, is calculated as follows:

For each of the 5 positions in the word, there are 10 possible choices for the letter (since repetition is allowed).

By the Fundamental Principle of Counting, the total number of 5-letter words with repetition allowed is the product of the number of choices for each position.

Next, we calculate the number of words that have no letter repeated. This means all 5 letters in the word are distinct. This is a permutation problem of selecting and arranging 5 distinct letters from the 10 available distinct letters.

The number of permutations of 10 different things taken 5 at a time is $P(10, 5)$.

The number of words which have at least one letter repeated is equal to the total number of words that can be formed (with repetition) minus the number of words that have no letter repeated.

The number of words which have at least one letter repeated is 69760.

The correct option is (A) 69760.

Given:

Total number of flags = $6$

The flags are of different colours (distinct).

Condition: Signals are sent by taking one or more flags at a time.

To Find:

The total number of signals that can be sent.

Solution:

A signal is formed by arranging one or more flags in a sequence. Since the flags are of different colours, the order of the flags matters, which means this is a permutation problem.

The possible number of flags taken at a time are 1, 2, 3, 4, 5, or 6.

We need to calculate the number of permutations of 6 different things taken $r$ at a time, for $r = 1, 2, 3, 4, 5, 6$, and sum these numbers up.

Number of signals using 1 flag = Permutations of 6 flags taken 1 at a time, $P(6, 1)$.

Number of signals using 2 flags = Permutations of 6 flags taken 2 at a time, $P(6, 2)$.

Number of signals using 3 flags = Permutations of 6 flags taken 3 at a time, $P(6, 3)$.

Number of signals using 4 flags = Permutations of 6 flags taken 4 at a time, $P(6, 4)$.

Number of signals using 5 flags = Permutations of 6 flags taken 5 at a time, $P(6, 5)$.

Number of signals using 6 flags = Permutations of 6 flags taken 6 at a time, $P(6, 6)$.

The total number of signals is the sum of the numbers of signals for each case (taking 1, 2, 3, 4, 5, or 6 flags).

The total number of signals that can be sent is 1956.

The correct option is (B) 1956.

Given:

Number of multiple choice questions = $3$

Number of choices for each question = $4$

To Find:

The number of ways in which a student can fail to get all answers correct.

Solution:

For each question, there are 4 independent choices. The total number of ways a student can answer the three questions is the product of the number of choices for each question.

There is only one way to get all answers correct (by selecting the correct option for each of the three questions). Let's assume there is a unique correct option for each question.

The number of ways a student can fail to get all answers correct is the total number of ways to answer minus the number of ways to get all answers correct.

The number of ways in which a student can fail to get all answers correct is 63.

The correct option is (D) 63.

Given:

Three parallel lines $l_1, l_2, l_3$ in the same plane.

Number of points on line $l_1 = m$

Number of points on line $l_2 = n$

Number of points on line $l_3 = k$

To Find:

The maximum number of triangles that can be formed with vertices chosen from these $m+n+k$ points.

Solution:

A triangle is formed by selecting three non-collinear points. The total number of points is $m + n + k$.

The total number of ways to choose any 3 points from these $m+n+k$ points, without considering collinearity, is given by the combination formula $\binom{m+n+k}{3}$.

However, this total includes cases where the three selected points are collinear. Since the lines $l_1, l_2, l_3$ are parallel, three points can be collinear only if all three points lie on the same line.

The number of ways to choose 3 points from line $l_1$ (which are collinear) is $\binom{m}{3}$. These combinations do not form a triangle.

The number of ways to choose 3 points from line $l_2$ (which are collinear) is $\binom{n}{3}$. These combinations do not form a triangle.

The number of ways to choose 3 points from line $l_3$ (which are collinear) is $\binom{k}{3}$. These combinations do not form a triangle.

The number of combinations of 3 points that are collinear is the sum of the ways to choose 3 points from each individual line.

The number of triangles formed is the total number of ways to choose 3 points minus the number of ways to choose 3 collinear points.

This formula represents the maximum number of triangles, as it excludes all sets of 3 points that lie on the same straight line.

Comparing this with the given options, the correct option is (B) $^{m + n + k}C_3 – ^mC_3 – ^nC_3 – ^kC_3$.

Given:

Total chairs numbered from 1 to 8 (8 chairs).

Number of women = 2

Number of men = 3

Women choose chairs from chairs 1 to 4 (4 chairs).

Men choose chairs from the remaining chairs.

To Find:

Total number of possible arrangements for the seating of 2 women and 3 men.

Solution:

The problem involves arranging distinct individuals (women and men) into distinct positions (numbered chairs). The order of selection and seating matters, so this is a permutation problem.

Step 1: Women choose chairs.

The 2 women choose and occupy 2 chairs from the 4 chairs numbered 1 to 4. The number of ways to select and arrange 2 women in 2 chairs out of 4 is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$, where $n=4$ (chairs) and $r=2$ (women).

Step 2: Men choose chairs.

After the 2 women have occupied their chairs, there are a total of $8 - 2 = 6$ chairs remaining. The 3 men must choose and occupy 3 chairs from these 6 remaining chairs.

The number of ways to select and arrange 3 men in 3 chairs out of the remaining 6 chairs is given by the permutation formula $P(n, r)$, where $n=6$ (remaining chairs) and $r=3$ (men).

Step 3: Total number of arrangements.

Since the seating of women and the seating of men are sequential and independent processes, the total number of possible arrangements is the product of the number of ways for each step, by the Fundamental Principle of Counting.

The total number of possible arrangements for the seating is $1440$.

The word is RACHIT.

The letters of the word RACHIT are R, A, C, H, I, T. There are 6 distinct letters.

Arranging these letters in alphabetical order, we get: A, C, H, I, R, T.

To find the rank of the word RACHIT in the dictionary, we list all the words alphabetically and count how many words appear before RACHIT.

We arrange the letters A, C, H, I, R, T in all possible $6!$ ways.

Calculate the number of words starting with letters that come before 'R' in the alphabetical order (A, C, H, I).

1. Words starting with A:

If a word starts with A, the remaining 5 letters (C, H, I, R, T) can be arranged in $5!$ ways.

Number of words starting with A $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

2. Words starting with C:

If a word starts with C, the remaining 5 letters (A, H, I, R, T) can be arranged in $5!$ ways.

Number of words starting with C $= 5! = 120$.

3. Words starting with H:

If a word starts with H, the remaining 5 letters (A, C, I, R, T) can be arranged in $5!$ ways.

Number of words starting with H $= 5! = 120$.

4. Words starting with I:

If a word starts with I, the remaining 5 letters (A, C, H, R, T) can be arranged in $5!$ ways.

Number of words starting with I $= 5! = 120$.

Total number of words starting with letters before R $= 120 + 120 + 120 + 120 = 4 \times 120 = 480$.

Now consider words starting with R. The second letter of RACHIT is A.

We look for words starting with R followed by a letter alphabetically before A from the remaining letters (A, C, H, I, T). There are no letters before A.

Consider words starting with RA. The remaining letters are C, H, I, T.

The third letter of RACHIT is C.

We look for words starting with RAC followed by a letter alphabetically before C from the remaining letters (C, H, I, T). There are no letters before C.

Consider words starting with RAC. The remaining letters are H, I, T.

The fourth letter of RACHIT is H.

We look for words starting with RACH followed by a letter alphabetically before H from the remaining letters (H, I, T). There are no letters before H.

Consider words starting with RACH. The remaining letters are I, T.

The fifth letter of RACHIT is I.

We look for words starting with RACHI followed by a letter alphabetically before I from the remaining letters (I, T). There are no letters before I.

Consider words starting with RACHI. The remaining letter is T.

The sixth letter of RACHIT is T.

The word RACHIT is the first word starting with RACHI followed by T.

Using the rank formula:

Alphabetical order of letters: A, C, H, I, R, T

Word: RACHIT

1. First letter R: Letters before R are A, C, H, I (4 letters). Contribution $= 4 \times 5! = 4 \times 120 = 480$. Remaining letters: A, C, H, I, T.

2. Second letter A: Letters before A in (A, C, H, I, T) are none (0 letters). Contribution $= 0 \times 4! = 0$. Remaining letters: C, H, I, T.

3. Third letter C: Letters before C in (C, H, I, T) are none (0 letters). Contribution $= 0 \times 3! = 0$. Remaining letters: H, I, T.

4. Fourth letter H: Letters before H in (H, I, T) are none (0 letters). Contribution $= 0 \times 2! = 0$. Remaining letters: I, T.

5. Fifth letter I: Letters before I in (I, T) are none (0 letters). Contribution $= 0 \times 1! = 0$. Remaining letter: T.

6. Sixth letter T: Letters before T in (T) are none (0 letters). Contribution $= 0 \times 0! = 0$.

Total number of words before RACHIT $= 480 + 0 + 0 + 0 + 0 + 0 = 480$.

The rank of the word is the number of words before it plus 1.

Rank of RACHIT $= 480 + 1 = 481$.

The rank of the word RACHIT is 481.

Given:

Total number of questions = 12.

The questions are divided into two groups, Group 1 and Group 2, each containing 6 questions.

Number of questions in Group 1 = 6.

Number of questions in Group 2 = 6.

A candidate must answer a total of 7 questions.

Constraint: The candidate is not permitted to attempt more than 5 questions from either group.

To Find:

The number of different ways of choosing 7 questions under the given constraints.

Solution:

Let $n_1$ be the number of questions chosen from Group 1 and $n_2$ be the number of questions chosen from Group 2.

The total number of questions to be answered is 7, so we must have $n_1 + n_2 = 7$.

Since there are 6 questions in each group, $0 \le n_1 \le 6$ and $0 \le n_2 \le 6$.

The constraint is that the candidate cannot attempt more than 5 questions from either group. This means $n_1 \le 5$ and $n_2 \le 5$.

We need to find pairs of integers $(n_1, n_2)$ such that:

$n_1 + n_2 = 7$

$n_1 \le 5$

$n_2 \le 5$

Given $n_1 + n_2 = 7$, the possible pairs $(n_1, n_2)$ are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).

Now, we apply the constraints $n_1 \le 5$ and $n_2 \le 5$ to these pairs:

• $(1, 6)$: Here $n_2 = 6$, which violates $n_2 \le 5$. So, this case is not allowed.
• $(2, 5)$: Here $n_1 = 2 \le 5$ and $n_2 = 5 \le 5$. This case is allowed.
• $(3, 4)$: Here $n_1 = 3 \le 5$ and $n_2 = 4 \le 5$. This case is allowed.
• $(4, 3)$: Here $n_1 = 4 \le 5$ and $n_2 = 3 \le 5$. This case is allowed.
• $(5, 2)$: Here $n_1 = 5 \le 5$ and $n_2 = 2 \le 5$. This case is allowed.
• $(6, 1)$: Here $n_1 = 6$, which violates $n_1 \le 5$. So, this case is not allowed.

The allowed combinations for $(n_1, n_2)$ are (2, 5), (3, 4), (4, 3), and (5, 2).

We calculate the number of ways for each allowed combination:

Case 1: Choosing 2 questions from Group 1 and 5 questions from Group 2 ($n_1=2, n_2=5$).

Number of ways = $\binom{6}{2} \times \binom{6}{5}$

$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$

$\binom{6}{5} = \frac{6!}{5!(6-5)!} = \frac{6!}{5!1!} = \frac{6}{1} = 6$

Number of ways for Case 1 = $15 \times 6 = 90$.

Case 2: Choosing 3 questions from Group 1 and 4 questions from Group 2 ($n_1=3, n_2=4$).

Number of ways = $\binom{6}{3} \times \binom{6}{4}$

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

$\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$

Number of ways for Case 2 = $20 \times 15 = 300$.

Case 3: Choosing 4 questions from Group 1 and 3 questions from Group 2 ($n_1=4, n_2=3$).

Number of ways = $\binom{6}{4} \times \binom{6}{3}$

$\binom{6}{4} = 15$

$\binom{6}{3} = 20$

Number of ways for Case 3 = $15 \times 20 = 300$.

Case 4: Choosing 5 questions from Group 1 and 2 questions from Group 2 ($n_1=5, n_2=2$).

Number of ways = $\binom{6}{5} \times \binom{6}{2}$

$\binom{6}{5} = 6$

$\binom{6}{2} = 15$

Number of ways for Case 4 = $6 \times 15 = 90$.

The total number of different ways of doing the questions is the sum of the ways for all allowed cases.

Total ways = Ways (Case 1) + Ways (Case 2) + Ways (Case 3) + Ways (Case 4)

Total ways = $90 + 300 + 300 + 90 = 780$.

The number of different ways of doing the questions is 780.

Given:

Total number of points in a plane = 18.

Number of collinear points among these 18 = 5.

No three points are collinear, except for these 5 points.

To Find:

The number of lines that can be formed by joining any two of these points.

Solution:

The number of lines that can be formed by joining any two points from $n$ distinct points is given by $\binom{n}{2}$, assuming no three points are collinear.

In this problem, we have a total of 18 points.

If no three of these 18 points were collinear, the total number of lines formed would be $\binom{18}{2}$.

However, there are 5 points which are collinear. These 5 points lie on a single straight line.

The number of lines that can be formed by choosing any two points from these 5 collinear points is $\binom{5}{2}$.

In the general calculation $\binom{18}{2}$, the lines formed by choosing any two of these 5 collinear points are counted as distinct lines. But these 5 points only form one line (the line on which they lie).

Therefore, we must subtract the number of lines formed by choosing any two of the 5 collinear points ($\binom{5}{2}$) from the total number of lines formed by 18 points ($\binom{18}{2}$), and then add back the single line formed by the 5 collinear points.

The number of lines is given by the formula:

Number of lines = $\binom{18}{2} - \binom{5}{2} + 1$

Calculate $\binom{18}{2}$:

$\binom{18}{2} = \frac{18!}{2!(18-2)!} = \frac{18!}{2!16!} = \frac{18 \times 17}{2 \times 1} = 9 \times 17 = 153$

Calculate $\binom{5}{2}$:

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$

Substitute these values into the formula:

Number of lines = $153 - 10 + 1$

Number of lines = $143 + 1 = 144$

The number of different lines that can be formed by joining the points is 144.

Given:

Total number of persons = 8.

Number of persons to be selected = 6.

Constraint: If person A is chosen, then person B must be chosen.

To Find:

The number of ways to make the selection under the given constraint.

Solution:

Let the set of 8 persons include specific persons A and B.

The constraint states that if A is selected for the group of 6, then B must also be selected.

We can consider the possible cases regarding the selection of persons A and B:

Case 1: Neither A nor B is selected.

In this case, both A and B are excluded from the selection. We need to choose 6 persons from the remaining $8 - 2 = 6$ persons.

The number of ways to choose 6 persons from the remaining 6 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Number of ways for Case 1 = $\binom{6}{6} = \frac{6!}{6!(6-6)!} = \frac{6!}{6!0!} = 1$.

Case 2: Both A and B are selected.

According to the constraint, if A is chosen, B must be chosen. This case includes the scenario where A is chosen, fulfilling the condition by also choosing B.

If both A and B are selected, we have already chosen 2 persons (A and B). We need to choose the remaining $6 - 2 = 4$ persons from the remaining $8 - 2 = 6$ persons (excluding A and B).

Number of ways for Case 2 = $\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5 \times 4!}{4! \times 2 \times 1} = \frac{6 \times 5}{2} = 15$.

Case 3: B is selected, but A is not selected.

This case is allowed by the constraint. The constraint "if A is chosen, then B must be chosen" does not impose any restriction when A is not chosen.

If B is selected and A is not selected, we have chosen B (1 person) and need to choose $6 - 1 = 5$ additional persons from the remaining $8 - 2 = 6$ persons (excluding A and B).

Number of ways for Case 3 = $\binom{6}{5} = \frac{6!}{5!(6-5)!} = \frac{6!}{5!1!} = \frac{6 \times 5!}{5! \times 1} = 6$.

Case 4: A is selected, but B is not selected.

This case is not allowed by the given constraint ("if the person A is chosen, then B must be chosen").

The number of ways for this case is 0.

The total number of ways to make the selection is the sum of the number of ways in the allowed cases (Case 1, Case 2, and Case 3).

Total number of ways = Ways (Case 1) + Ways (Case 2) + Ways (Case 3)

Total number of ways $= 1 + 15 + 6 = 22$.

Alternate Solution:

We can also solve this problem by considering the total number of unrestricted selections and subtracting the number of forbidden selections.

The total number of ways to select 6 persons from 8 without any constraint is $\binom{8}{6}$.

Total unrestricted ways $= \binom{8}{6} = \frac{8!}{6!(8-6)!} = \frac{8!}{6!2!} = \frac{8 \times 7 \times 6!}{6! \times 2 \times 1} = \frac{8 \times 7}{2} = 28$.

The constraint is violated only when person A is chosen, but person B is not chosen.

Let's find the number of ways where A is selected and B is not selected for the group of 6.

If A is selected and B is not selected, we must include A in the selection (1 person), exclude B, and then choose the remaining $6 - 1 = 5$ persons from the remaining $8 - 2 = 6$ persons (excluding A and B).

Number of ways where A is selected but B is not selected = $\binom{1}{1} \times \binom{1}{0} \times \binom{6}{5}$.

$\binom{1}{1} = 1$ (Choose A)

$\binom{1}{0} = 1$ (Do not choose B)

$\binom{6}{5} = 6$ (Choose 5 from the remaining 6)

Number of forbidden ways $= 1 \times 1 \times 6 = 6$.

The number of ways satisfying the constraint is the total number of unrestricted ways minus the number of forbidden ways.

Number of ways with constraint = Total unrestricted ways - Number of forbidden ways

Number of ways with constraint $= 28 - 6 = 22$.

Both methods yield the same result.

The number of ways selections can be made is 22.

Given:

Total number of persons = 12.

We need to select a committee of 5 persons.

The committee must have a chairperson.

To Find:

The number of ways to select a committee of five persons with a chairperson from 12 persons.

Solution:

We need to select a committee of 5 persons from 12, and one of the selected persons must be designated as the chairperson. This selection involves two steps: first choosing the members of the committee, and then selecting one of them to be the chairperson OR first selecting the chairperson and then selecting the remaining members.

We will use the second approach, which is also suggested by the hint.

Step 1: Select the chairperson.

The chairperson can be any one of the 12 persons. The number of ways to select 1 chairperson from 12 persons is given by $\binom{12}{1}$.

Number of ways to select the chairperson $= \binom{12}{1} = \frac{12!}{1!(12-1)!} = \frac{12!}{1!11!} = \frac{12 \times 11!}{1 \times 11!} = 12$.

Step 2: Select the remaining committee members.

After selecting the chairperson, there are $12 - 1 = 11$ persons remaining.

The committee needs a total of 5 persons. Since the chairperson has already been selected, we need to select the remaining $5 - 1 = 4$ members for the committee from the remaining 11 persons.

The number of ways to select 4 members from 11 persons is given by $\binom{11}{4}$.

Number of ways to select the remaining 4 members $= \binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!} = \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}$.

Calculating the value:

$\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = \frac{11 \times (5 \times 2) \times (3 \times 3) \times (4 \times 2)}{24}$.

We can simplify the calculation:

$\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{24} = 11 \times \frac{10 \times 9 \times 8}{24} = 11 \times \frac{720}{24} = 11 \times 30 = 330$.

Total number of ways:

To find the total number of ways to select a committee of five persons with a chairperson, we multiply the number of ways to perform Step 1 and Step 2.

Total number of ways = (Number of ways to select chairperson) $\times$ (Number of ways to select remaining members)

Total number of ways $= \binom{12}{1} \times \binom{11}{4} = 12 \times 330$.

$12 \times 330 = 3960$.

Alternate Solution:

We can also solve this by first selecting the 5 committee members and then choosing one of them as the chairperson.

Step 1: Select the 5 committee members.

The number of ways to select 5 persons from 12 is $\binom{12}{5}$.

Number of ways to select 5 members $= \binom{12}{5} = \frac{12!}{5!(12-5)!} = \frac{12!}{5!7!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}$.

Calculating the value:

$\binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{120}$.

$\frac{12 \times 10}{120} = \frac{120}{120} = 1$.

So, $\binom{12}{5} = 11 \times 9 \times 8 = 11 \times 72 = 792$.

Step 2: Select the chairperson from the selected committee members.

From the 5 selected committee members, we need to choose 1 person to be the chairperson. The number of ways to do this is $\binom{5}{1}$.

Number of ways to select the chairperson $= \binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = \frac{5 \times 4!}{1 \times 4!} = 5$.

Total number of ways:

Total number of ways = (Number of ways to select 5 members) $\times$ (Number of ways to select chairperson from them)

Total number of ways $= \binom{12}{5} \times \binom{5}{1} = 792 \times 5$.

$792 \times 5 = 3960$.

Both methods yield the same result.

The number of ways a committee of five persons with a chairperson can be selected from 12 persons is 3960.

Given:

Each automobile license plate consists of two different letters followed by three different digits.

There are 26 letters in the English alphabet.

There are 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).

To Find:

The total number of different automobile license plates that can be made.

Solution:

The formation of a license plate involves selecting and arranging letters and digits according to the given rules.

The license plate has two parts: a letter part (first two positions) and a digit part (last three positions).

Let's consider the letter part first. We need to choose 2 different letters from 26 available letters and arrange them in the first two positions. The number of ways to do this is the number of permutations of 26 items taken 2 at a time, denoted as $P(26, 2)$.

The formula for permutations is $P(n, k) = \frac{n!}{(n-k)!}$.

Number of ways to choose and arrange the two different letters $= P(26, 2) = \frac{26!}{(26-2)!} = \frac{26!}{24!} = 26 \times 25$.

$26 \times 25 = 650$.

Now, let's consider the digit part. We need to choose 3 different digits from 10 available digits (0 through 9) and arrange them in the last three positions. The number of ways to do this is the number of permutations of 10 items taken 3 at a time, denoted as $P(10, 3)$.

Number of ways to choose and arrange the three different digits $= P(10, 3) = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 10 \times 9 \times 8$.

$10 \times 9 \times 8 = 720$.

Since the selection and arrangement of letters and digits are independent events, the total number of different license plates is the product of the number of ways to form the letter part and the number of ways to form the digit part.

Total number of license plates = (Number of ways for letters) $\times$ (Number of ways for digits)

Total number of license plates $= P(26, 2) \times P(10, 3) = 650 \times 720$.

Calculating the product:

$650 \times 720 = 65 \times 10 \times 72 \times 10 = 65 \times 72 \times 100$.

$65 \times 72 = 4680$.

Total number of license plates $= 4680 \times 100 = 468000$.

The number of different automobile license plates that can be made is 468,000.

Given:

Number of black balls in the bag = 5.

Number of red balls in the bag = 6.

To Find:

The number of ways to select 2 black balls and 3 red balls from the lot.

Solution:

We need to perform two independent selections:

1. Select 2 black balls from the 5 black balls.

2. Select 3 red balls from the 6 red balls.

The number of ways to select $k$ items from a set of $n$ distinct items, without regard to the order of selection, is given by the combination formula $\binom{n}{k}$.

Number of ways to select 2 black balls from 5 is $\binom{5}{2}$.

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{5 \times 4}{2} = \frac{20}{2} = 10$.

Number of ways to select 3 red balls from 6 is $\binom{6}{3}$.

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} = \frac{6 \times 5 \times 4}{6} = \frac{120}{6} = 20$.

Since the selection of black balls and the selection of red balls are independent events, the total number of ways to select 2 black balls and 3 red balls is the product of the number of ways for each selection.

Total number of ways = (Ways to select 2 black balls) $\times$ (Ways to select 3 red balls)

Total number of ways $= \binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200$.

The number of ways in which 2 black and 3 red balls can be selected from the lot is 200.

Given:

Total number of distinct things = $n$.

Number of things taken together for permutation = $r$.

Constraint: 3 particular things must occur together in the permutation.

To Find:

The number of permutations of $n$ distinct things taken $r$ together, such that 3 particular things always occur together.

Solution:

Let the 3 particular things be denoted as $p_1, p_2, p_3$. Since these 3 particular things must occur in the permutation of length $r$, they are included in the set of $r$ things being permuted.

For a permutation of length $r$ to contain these 3 particular things, the other $r-3$ things must be selected from the remaining $n-3$ distinct things (total $n$ things minus the 3 particular ones).

The number of ways to select the remaining $r-3$ things from the $n-3$ things is given by the combination formula $\binom{n-3}{r-3}$. This is possible only if $n-3 \ge r-3$, which means $n \ge r$, and $r-3 \ge 0$, which means $r \ge 3$. If $r < 3$ or $r > n$, the number of such permutations is 0.

Number of ways to select the remaining $r-3$ things $= \binom{n-3}{r-3}$.

Now, we have a set of $r$ things: the 3 particular things $\{p_1, p_2, p_3\}$ and the $r-3$ selected things from the rest. We need to arrange these $r$ things such that the 3 particular things always occur together.

Consider the group of 3 particular things $\{p_1, p_2, p_3\}$ as a single block or unit. Within this block, the 3 particular things can be arranged among themselves in $3!$ ways.

Number of ways to arrange the 3 particular things within their block $= 3! = 3 \times 2 \times 1 = 6$.

Now, we are arranging the block of 3 particular things and the $r-3$ other selected things. This is equivalent to arranging $1 + (r-3) = r-2$ distinct units (the block is one unit, and each of the $r-3$ selected things is a unit).

The number of ways to arrange these $r-2$ units is $(r-2)!$.

Number of ways to arrange the $r-2$ units $= (r-2)!$.

To find the total number of permutations, we multiply the number of ways for each independent step:

1. Number of ways to select the $r-3$ items from $n-3$: $\binom{n-3}{r-3}$.

2. Number of ways to arrange the 3 particular items within their block: $3!$.

3. Number of ways to arrange the block and the $r-3$ selected items as $r-2$ units: $(r-2)!$.

Total number of permutations = (Ways to select $r-3$) $\times$ (Ways to arrange 3 within block) $\times$ (Ways to arrange $r-2$ units)

Total number of permutations $= \binom{n-3}{r-3} \times 3! \times (r-2)!$

We can expand the combination term:

$\binom{n-3}{r-3} = \frac{(n-3)!}{(r-3)!( (n-3) - (r-3) )!} = \frac{(n-3)!}{(r-3)!(n-r)!}$

Substituting this into the formula:

Total number of permutations $= \frac{(n-3)!}{(r-3)!(n-r)!} \times 3! \times (r-2)!$

This formula is valid for $n \ge r \ge 3$. If $r < 3$ or $r > n$, the number of permutations is 0.

Alternatively, we can write the number of permutations as:

Total number of permutations $= P(n-3, r-3) \times 3! \times (r-2)!$

where $P(n-3, r-3) = \frac{(n-3)!}{(n-3-(r-3))!} = \frac{(n-3)!}{(n-r)!}$ is the number of permutations of $n-3$ things taken $r-3$ at a time.

The number of permutations is $\binom{n-3}{r-3} \times (r-2)! \times 3!$.

Given:

The word is ‘TRIANGLE’.

The letters are T, R, I, A, N, G, L, E.

Total number of letters = 8.

To Find:

The number of different words that can be formed from the letters of the word ‘TRIANGLE’ such that no two vowels are together.

Solution:

First, identify the vowels and consonants in the word ‘TRIANGLE’.

Vowels: I, A, E (There are 3 vowels).

Consonants: T, R, N, G, L (There are 5 consonants).

Total letters = 3 vowels + 5 consonants = 8.

To ensure that no two vowels are together, we can first arrange the consonants and then place the vowels in the spaces created between the consonants and at the ends.

Step 1: Arrange the consonants.

There are 5 distinct consonants (T, R, N, G, L). The number of ways to arrange these 5 consonants is $5!$.

Number of ways to arrange consonants $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Step 2: Create spaces for the vowels.

When we arrange the 5 consonants, there are 6 possible positions where the vowels can be placed so that no two vowels are adjacent. Let 'C' represent a consonant and '\_' represent a space.

The arrangement of consonants creates the following spaces:

$\_ C \_ C \_ C \_ C \_ C \_$

There are 6 available spaces indicated by '\_'.

Step 3: Place the vowels in the spaces.

We have 3 distinct vowels (I, A, E) to place in the 6 available spaces. Since the vowels are distinct, we need to select 3 of the 6 spaces and arrange the 3 vowels in those selected spaces.

The number of ways to choose 3 spaces from 6 and arrange the 3 distinct vowels in those spaces is given by the number of permutations of 6 items taken 3 at a time, denoted as $P(6, 3)$.

Number of ways to place vowels $= P(6, 3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}$.

$P(6, 3) = \frac{6 \times 5 \times 4 \times 3!}{3!} = 6 \times 5 \times 4 = 120$.

Step 4: Calculate the total number of words.

The total number of different words formed with no two vowels together is the product of the number of ways to arrange the consonants and the number of ways to place the vowels in the available spaces.

Total number of words = (Number of ways to arrange consonants) $\times$ (Number of ways to place vowels)

Total number of words $= 5! \times P(6, 3) = 120 \times 120$.

$120 \times 120 = 14400$.

The number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together is 14,400.

Given:

We need to find the number of positive integers.

The integers must be greater than 6000 and less than 7000.

The integers must be divisible by 5.

No digit is to be repeated in the integer.

To Find:

The number of such integers.

Solution:

The integers must be greater than 6000 and less than 7000. This means they are 4-digit numbers.

Let the 4-digit number be represented as $d_1 d_2 d_3 d_4$, where $d_1, d_2, d_3, d_4$ are the digits.

Since the number is greater than 6000 and less than 7000, the first digit $d_1$ must be 6.

The number must be divisible by 5. A number is divisible by 5 if its last digit $d_4$ is either 0 or 5.

The digits available are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

No digit can be repeated.

We consider the cases based on the last digit, $d_4$.

Case 1: The last digit $d_4$ is 0.

The first digit $d_1 = 6$.

The last digit $d_4 = 0$.

The digits used so far are 6 and 0. Since repetition is not allowed, these two digits cannot be used for $d_2$ and $d_3$.

The remaining available digits are the 10 original digits excluding 6 and 0, which are {1, 2, 3, 4, 5, 7, 8, 9}. There are 8 remaining digits.

We need to select and place 2 distinct digits for $d_2$ and $d_3$ from these 8 remaining digits.

The number of choices for $d_2$ is 8 (any of the remaining digits).

After choosing $d_2$, the number of choices for $d_3$ is 7 (any of the remaining digits except $d_2$).

The number of ways to choose and place $d_2$ and $d_3$ is $8 \times 7 = 56$. This is also the number of permutations of 8 distinct things taken 2 at a time, $P(8, 2)$.

$P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!} = 8 \times 7 = 56$.

The number of such integers in Case 1 is the number of ways to fix $d_1$ and $d_4$ times the number of ways to choose and place $d_2$ and $d_3$.

Number of integers in Case 1 $= 1 \times 56 \times 1 = 56$.

Case 2: The last digit $d_4$ is 5.

The first digit $d_1 = 6.

The last digit $d_4 = 5$.

The digits used so far are 6 and 5. Since repetition is not allowed, these two digits cannot be used for $d_2$ and $d_3$.

The remaining available digits are the 10 original digits excluding 6 and 5, which are {0, 1, 2, 3, 4, 7, 8, 9}. There are 8 remaining digits.

We need to select and place 2 distinct digits for $d_2$ and $d_3$ from these 8 remaining digits.

The number of choices for $d_2$ is 8 (any of the remaining digits).

After choosing $d_2$, the number of choices for $d_3$ is 7 (any of the remaining digits except $d_2$).

The number of ways to choose and place $d_2$ and $d_3$ is $8 \times 7 = 56$. This is $P(8, 2)$.

$P(8, 2) = 56$.

The number of such integers in Case 2 is the number of ways to fix $d_1$ and $d_4$ times the number of ways to choose and place $d_2$ and $d_3$.

Number of integers in Case 2 $= 1 \times 56 \times 1 = 56$.

The total number of positive integers greater than 6000 and less than 7000 which are divisible by 5 and have no repeated digits is the sum of the numbers of integers in Case 1 and Case 2.

Total number of integers = Number of integers in Case 1 + Number of integers in Case 2

Total number of integers $= 56 + 56 = 112$.

The number of such integers is 112.

Given:

Total number of persons = 10 (P₁, P₂, ..., P₁₀).

Number of persons to be arranged in a line = 5.

Constraint: In each arrangement, P₁ must occur, and P₄ and P₅ must not occur.

To Find:

The number of such possible arrangements.

Solution:

We need to form a linear arrangement of 5 persons with the given conditions.

The conditions are:

1. The arrangement must include P₁.

2. The arrangement must not include P₄ and P₅.

Since P₁ must be in the arrangement, we can consider P₁ as already selected for the group of 5.

Since P₄ and P₅ must not occur, they are excluded from the set of available persons.

The original set of 10 persons is {P₁, P₂, P₃, P₄, P₅, P₆, P₇, P₈, P₉, P₁₀}.

After excluding P₄ and P₅, the remaining persons are {P₁, P₂, P₃, P₆, P₇, P₈, P₉, P₁₀}. There are $10 - 2 = 8$ persons available.

From this set of 8 persons, we must select P₁. So, P₁ is one of the 5 persons in the arrangement.

We need to select the remaining $5 - 1 = 4$ persons from the set of available persons, excluding P₁, P₄, and P₅. The set of persons from which we choose the remaining 4 is {P₂, P₃, P₆, P₇, P₈, P₉, P₁₀}. There are $8 - 1 = 7$ persons in this set.

Step 1: Select the remaining 4 persons for the group of 5.

We need to select 4 persons from the 7 available persons (excluding P₁, P₄, P₅). The number of ways to do this is given by $\binom{7}{4}$.

Number of ways to select the remaining 4 persons $= \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} = \frac{7 \times 6 \times 5}{6} = 7 \times 5 = 35$.

After this step, we have selected a group of 5 persons, which includes P₁ and 4 others chosen from {P₂, P₃, P₆, P₇, P₈, P₉, P₁₀}.

Step 2: Arrange the selected 5 persons in a line.

For each group of 5 selected persons, these 5 distinct persons can be arranged in a line in $5!$ ways.

Number of ways to arrange the 5 selected persons $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

To find the total number of possible arrangements satisfying the conditions, we multiply the number of ways to perform Step 1 and Step 2.

Total number of arrangements = (Number of ways to select the group of 5) $\times$ (Number of ways to arrange the group of 5)

Total number of arrangements $= \binom{7}{4} \times 5! = 35 \times 120$.

$35 \times 120 = 35 \times 12 \times 10 = (35 \times 10 + 35 \times 2) \times 10 = (350 + 70) \times 10 = 420 \times 10 = 4200$.

The number of such possible arrangements is 4200.

Given:

Total number of lamps in a hall = 10.

Each lamp can be switched on independently.

To Find:

The number of ways in which the hall can be illuminated.

Solution:

Each lamp in the hall can be in one of two states: either it is switched ON or it is switched OFF.

Since there are 10 lamps, and each lamp can be switched on or off independently, the total number of possible states for all 10 lamps is the product of the number of states for each lamp.

For the first lamp, there are 2 possibilities (ON or OFF).

For the second lamp, there are 2 possibilities (ON or OFF).

... and so on, for all 10 lamps.

The total number of ways the lamps can be switched ON or OFF is $2 \times 2 \times 2 \times ...$ (10 times).

Total number of possible states = $2^{10}$.

$2^{10} = 1024$.

The hall is illuminated if at least one lamp is switched ON.

The total number of possible states ($2^{10}$) includes one specific state where all the lamps are switched OFF. In this state, the hall is not illuminated.

To find the number of ways the hall can be illuminated, we need to exclude the case where all lamps are OFF from the total number of possible states.

Number of ways to illuminate the hall = (Total number of possible states) - (Number of ways all lamps are OFF)

Number of ways all lamps are OFF = 1 (This occurs only when each of the 10 lamps is switched OFF).

Number of ways to illuminate the hall = $2^{10} - 1$.

$2^{10} - 1 = 1024 - 1 = 1023$.

The number of ways in which the hall can be illuminated is 1023.

Given:

A box contains:

• White balls: 2
• Black balls: 3
• Red balls: 4

Total number of balls $= 2 + 3 + 4 = 9$.

We need to draw 3 balls from the box.

Constraint: At least one black ball must be included in the draw.

To Find:

The number of ways to draw 3 balls such that at least one black ball is included.

Solution:

The condition "at least one black ball" means that in the group of 3 balls drawn, there can be exactly 1 black ball, exactly 2 black balls, or exactly 3 black balls.

Let's consider each case:

Case 1: Exactly 1 black ball is drawn.

If exactly 1 black ball is drawn, the remaining $3 - 1 = 2$ balls must be drawn from the non-black balls. The non-black balls are the white and red balls, totaling $2 + 4 = 6$ balls.

Number of ways to choose 1 black ball from 3 $= \binom{3}{1}$.

Number of ways to choose 2 non-black balls from 6 $= \binom{6}{2}$.

Number of ways for Case 1 $= \binom{3}{1} \times \binom{6}{2}$.

$\binom{3}{1} = 3$

$\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$

Number of ways for Case 1 $= 3 \times 15 = 45$.

Case 2: Exactly 2 black balls are drawn.

If exactly 2 black balls are drawn, the remaining $3 - 2 = 1$ ball must be drawn from the non-black balls (6 non-black balls).

Number of ways to choose 2 black balls from 3 $= \binom{3}{2}$.

Number of ways to choose 1 non-black ball from 6 $= \binom{6}{1}$.

Number of ways for Case 2 $= \binom{3}{2} \times \binom{6}{1}$.

$\binom{3}{2} = \frac{3 \times 2}{2 \times 1} = 3$

$\binom{6}{1} = 6$

Number of ways for Case 2 $= 3 \times 6 = 18$.

Case 3: Exactly 3 black balls are drawn.

If exactly 3 black balls are drawn, the remaining $3 - 3 = 0$ balls must be drawn from the non-black balls (6 non-black balls).

Number of ways to choose 3 black balls from 3 $= \binom{3}{3}$.

Number of ways to choose 0 non-black balls from 6 $= \binom{6}{0}$.

Number of ways for Case 3 $= \binom{3}{3} \times \binom{6}{0}$.

$\binom{3}{3} = 1$

$\binom{6}{0} = 1$

Number of ways for Case 3 $= 1 \times 1 = 1$.

The total number of ways to draw 3 balls with at least one black ball is the sum of the ways in the three cases:

Total ways = Ways (Case 1) + Ways (Case 2) + Ways (Case 3)

Total ways $= 45 + 18 + 1 = 64$.

Alternate Solution:

We can find the total number of ways to draw 3 balls from the 9 balls without any restriction, and then subtract the number of ways to draw 3 balls with no black ball.

Total number of ways to draw 3 balls from 9 $= \binom{9}{3}$.

$\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$.

Number of ways to draw 3 balls with no black ball: This means we draw 3 balls from the non-black balls (2 white + 4 red = 6 balls).

Number of ways to choose 3 non-black balls from 6 $= \binom{6}{3}$.

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20$.

Number of ways to draw at least one black ball = (Total ways to draw 3 balls) - (Ways to draw 3 balls with no black ball)

Number of ways $= 84 - 20 = 64$.

Both methods yield the same result.

The number of ways three balls can be drawn from the box, if at least one black ball is to be included, is 64.

Note: The hint provided in the question, ³C₁ × ⁶C₂ + ³C₂ × ⁶C₂ + ³C₃, seems to have a typo in the second term. The correct calculation for the case of exactly 2 black balls requires choosing 1 non-black ball from 6, which is ⁶C₁, not ⁶C₂.

Given:

We need to find the value of $\binom{r}{2}$.

To Find:

$\binom{r}{2}$.

Solution:

We will use the property of combinations relating consecutive terms:

$\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$

Consider the ratio of equation (ii) to equation (i):

$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{84}{36}$

Using the property with $k=r$: $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$

So, $\frac{n-r+1}{r} = \frac{84}{36}$

Simplify the fraction $\frac{84}{36}$: Divide both numerator and denominator by their greatest common divisor, which is 12.

$\frac{84}{36} = \frac{84 \div 12}{36 \div 12} = \frac{7}{3}$

Thus, we have:

Cross-multiply:

$3(n-r+1) = 7r$

$3n - 3r + 3 = 7r$

$3n + 3 = 10r$

Now, consider the ratio of equation (iii) to equation (ii):

$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{126}{84}$

Using the property with $k=r+1$: $\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-(r+1)+1}{r+1} = \frac{n-r-1+1}{r+1} = \frac{n-r}{r+1}$

So, $\frac{n-r}{r+1} = \frac{126}{84}$

Simplify the fraction $\frac{126}{84}$: Divide both numerator and denominator by their greatest common divisor, which is 42.

$\frac{126}{84} = \frac{126 \div 42}{84 \div 42} = \frac{3}{2}$

Thus, we have:

Cross-multiply:

$2(n-r) = 3(r+1)$

$2n - 2r = 3r + 3$

Now we have a system of two linear equations in two variables $n$ and $r$ (from equations (v) and (vii)):

Equation (v): $3n = 10r - 3$

Equation (vii): $2n = 5r + 3$

We can solve this system. Multiply equation (vii) by 3 and equation (v) by 2 to eliminate $n$ (or multiply equation (vii) by 2 to easily eliminate $r$ after rearranging equation v):

From (v): $3n - 10r = -3$

From (vii): $2n - 5r = 3$

Multiply (vii) by 2: $4n - 10r = 6$

Now subtract the modified equation (v) from the modified equation (vii):

$(4n - 10r) - (3n - 10r) = 6 - (-3)$

$4n - 10r - 3n + 10r = 6 + 3$

$n = 9$

Now substitute the value of $n=9$ into either equation (v) or (vii) to find $r$. Using equation (vii):

$2(9) = 5r + 3$

$18 = 5r + 3$

$18 - 3 = 5r$

$15 = 5r$

$r = \frac{15}{5} = 3$

So, $n=9$ and $r=3$.

We can verify these values with the given combinations:

$\binom{9}{3-1} = \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36$ (Matches given value)

$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$ (Matches given value)

$\binom{9}{3+1} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = \frac{3024}{24} = 126$ (Matches given value)

The values of $n=9$ and $r=3$ are correct.

We need to find $\binom{r}{2}$. Since $r=3$, we need to find $\binom{3}{2}$.

$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2!}{2! \times 1} = 3$.

The value of $\binom{r}{2}$ is 3.

Given:

Set of distinct digits = {3, 5, 7, 8, 9}.

We need to form integers greater than 7000.

No digit is to be repeated.

To Find:

The number of integers greater than 7000 that can be formed using these digits without repetition.

Solution:

The integers greater than 7000 that can be formed using the digits {3, 5, 7, 8, 9} without repetition can be either 4-digit numbers or 5-digit numbers.

Case 1: 4-digit numbers.

A 4-digit number formed using these digits will be greater than 7000 if its first digit is 7, 8, or 9.

Let the 4-digit number be $d_1 d_2 d_3 d_4$.

The first digit, $d_1$, must be from the set {7, 8, 9}. There are 3 choices for the first digit.

After choosing the first digit, there are $5 - 1 = 4$ digits remaining from the original set.

The remaining three digits ($d_2, d_3, d_4$) must be chosen from these 4 remaining digits and arranged in the remaining 3 positions.

The number of ways to choose and arrange 3 distinct digits from 4 available digits is given by the number of permutations $P(4, 3)$.

Number of ways to fill the remaining 3 positions = $P(4, 3) = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \times 3 \times 2 = 24$.

The number of 4-digit integers greater than 7000 is the product of the number of choices for the first digit and the number of ways to arrange the remaining three digits.

Number of 4-digit integers $> 7000 = (\text{Choices for } d_1) \times (\text{Ways to arrange } d_2 d_3 d_4)$

Number of 4-digit integers $> 7000 = 3 \times 24 = 72$.

Case 2: 5-digit numbers.

Since we are using only the digits {3, 5, 7, 8, 9}, any 5-digit number formed using these distinct digits will be greater than 7000 (the smallest 5-digit number would start with 3 followed by other digits, like 35789, which is greater than 7000).

We use all 5 distinct digits to form a 5-digit number.

The number of ways to arrange 5 distinct digits is given by $5!$.

Number of 5-digit integers = $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

The total number of integers greater than 7000 that can be formed with the given digits without repetition is the sum of the numbers from Case 1 and Case 2.

Total number of integers = (Number of 4-digit integers $> 7000$) + (Number of 5-digit integers)

Total number of integers $= 72 + 120 = 192$.

The number of such integers is 192.

Given:

Number of lines drawn in a plane = 20.

Condition 1: No two lines are parallel.

Condition 2: No three lines are concurrent (intersect at the same point).

To Find:

The number of points at which these 20 lines intersect each other.

Solution:

When two distinct lines in a plane are not parallel, they intersect at exactly one point.

The problem states that no two of the 20 lines are parallel, which means every pair of lines will intersect at a unique point.

The problem also states that no three lines are concurrent. This means that the intersection point of any two lines is distinct from the intersection point of any other pair of lines.

Therefore, each intersection point corresponds to a unique pair of lines.

To find the total number of intersection points, we need to find the number of ways to choose any two distinct lines from the 20 lines.

The number of ways to choose $k$ items from a set of $n$ distinct items, without regard to the order of selection, is given by the combination formula $\binom{n}{k}$.

In this case, we are choosing 2 lines from the 20 available lines.

Number of ways to choose 2 lines from 20 $= \binom{20}{2}$.

Calculate $\binom{20}{2}$:

$\binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20!}{2!18!} = \frac{20 \times 19 \times 18!}{2 \times 1 \times 18!} = \frac{20 \times 19}{2}$.

$\frac{20 \times 19}{2} = 10 \times 19 = 190$.

Since each pair of lines intersects at a unique point and all these intersection points are distinct (due to the non-concurrent condition), the total number of intersection points is equal to the number of ways to choose 2 lines from 20.

The number of points at which the 20 lines will intersect each other is 190.

Given:

All telephone numbers have six digits.

The first two digits are restricted to one of the following pairs: 41, 42, 46, 62, or 64.

All six digits in the telephone number must be distinct.

The digits available are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. There are 10 distinct digits.

To Find:

The number of such telephone numbers with distinct digits.

Solution:

Let the six-digit telephone number be represented as $d_1 d_2 d_3 d_4 d_5 d_6$.

The first two digits ($d_1 d_2$) can be any of the 5 given pairs: 41, 42, 46, 62, or 64.

For the digits to be distinct throughout the six positions, the digits used for the first two positions cannot be used for the remaining four positions ($d_3, d_4, d_5, d_6$).

Let's consider each allowed pair for the first two digits and calculate the number of ways to fill the remaining four positions with distinct digits.

Case 1: The first two digits are 41.

$d_1 = 4$, $d_2 = 1$.

The digits 4 and 1 have been used. The remaining available digits are the 10 original digits excluding 4 and 1, which are {0, 2, 3, 5, 6, 7, 8, 9}. There are $10 - 2 = 8$ remaining digits.

We need to choose and arrange 4 distinct digits for the positions $d_3, d_4, d_5, d_6$ from these 8 remaining digits.

The number of ways to do this is the number of permutations of 8 items taken 4 at a time, $P(8, 4)$.

$P(8, 4) = \frac{8!}{(8-4)!} = \frac{8!}{4!} = 8 \times 7 \times 6 \times 5$.

$8 \times 7 \times 6 \times 5 = 56 \times 30 = 1680$.

Number of telephone numbers starting with 41 with distinct digits = 1680.

Case 2: The first two digits are 42.

$d_1 = 4$, $d_2 = 2$.

The digits 4 and 2 have been used. The remaining available digits are {0, 1, 3, 5, 6, 7, 8, 9}. There are 8 remaining digits.

We need to arrange 4 distinct digits in the remaining 4 positions from these 8 digits. The number of ways is $P(8, 4) = 1680$.

Number of telephone numbers starting with 42 with distinct digits = 1680.

Case 3: The first two digits are 46.

$d_1 = 4$, $d_2 = 6$.

The digits 4 and 6 have been used. The remaining available digits are {0, 1, 2, 3, 5, 7, 8, 9}. There are 8 remaining digits.

We need to arrange 4 distinct digits in the remaining 4 positions from these 8 digits. The number of ways is $P(8, 4) = 1680$.

Number of telephone numbers starting with 46 with distinct digits = 1680.

Case 4: The first two digits are 62.

$d_1 = 6$, $d_2 = 2$.

The digits 6 and 2 have been used. The remaining available digits are {0, 1, 3, 4, 5, 7, 8, 9}. There are 8 remaining digits.

We need to arrange 4 distinct digits in the remaining 4 positions from these 8 digits. The number of ways is $P(8, 4) = 1680$.

Number of telephone numbers starting with 62 with distinct digits = 1680.

Case 5: The first two digits are 64.

$d_1 = 6$, $d_2 = 4$.

The digits 6 and 4 have been used. The remaining available digits are {0, 1, 2, 3, 5, 7, 8, 9}. There are 8 remaining digits.

We need to arrange 4 distinct digits in the remaining 4 positions from these 8 digits. The number of ways is $P(8, 4) = 1680$.

Number of telephone numbers starting with 64 with distinct digits = 1680.

The total number of telephone numbers with all six digits distinct is the sum of the numbers from all these cases.

Total number of telephone numbers = (Ways for 41) + (Ways for 42) + (Ways for 46) + (Ways for 62) + (Ways for 64)

Total number of telephone numbers $= 1680 + 1680 + 1680 + 1680 + 1680 = 5 \times 1680$.

$5 \times 1680 = 8400$.

The number of telephone numbers that have all six digits distinct is 8400.

Given:

Total number of questions in the examination = 5.

Number of questions the student has to answer = 4.

Questions 1 and 2 are compulsory.

To Find:

The number of ways in which the student can choose the questions to answer.

Solution:

The student must answer a total of 4 questions.

Questions 1 and 2 are compulsory, meaning the student must answer these two questions.

So, out of the 4 questions the student has to answer, 2 are already determined (Question 1 and Question 2).

The remaining number of questions the student needs to choose is $4 - 2 = 2$.

These remaining 2 questions must be chosen from the questions that are not compulsory.

The total number of questions is 5. The compulsory questions are 1 and 2. So, the non-compulsory questions are 3, 4, and 5.

The number of non-compulsory questions = $5 - 2 = 3$.

The student needs to choose 2 questions from these 3 non-compulsory questions.

The number of ways to choose 2 questions from 3 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Number of ways to choose the remaining 2 questions $= \binom{3}{2}$.

$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2!}{2! \times 1} = 3$.

Since the compulsory questions are fixed, the total number of ways the student can make the choice is simply the number of ways to choose the remaining questions from the non-compulsory ones.

Total number of ways = Number of ways to choose 2 questions from the 3 non-compulsory questions.

Total number of ways $= \binom{3}{2} = 3$.

The possible choices for the remaining 2 questions are:

• Question 3 and Question 4
• Question 3 and Question 5
• Question 4 and Question 5

The complete sets of questions answered would be:

• {Question 1, Question 2, Question 3, Question 4}
• {Question 1, Question 2, Question 3, Question 5}
• {Question 1, Question 2, Question 4, Question 5}

The number of ways in which the student can make the choice is 3.

Given:

A convex polygon has 44 diagonals.

To Find:

The number of sides of the polygon.

Solution:

Let the number of sides of the convex polygon be $n$.

A diagonal of a polygon is a line segment connecting two non-adjacent vertices.

The total number of line segments that can be formed by connecting any two vertices of an $n$-sided polygon is the number of ways to choose 2 vertices from $n$ vertices, which is given by the combination formula $\binom{n}{2}$.

$\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2}$.

These line segments are either sides of the polygon or diagonals of the polygon.

An $n$-sided polygon has exactly $n$ sides.

The number of diagonals is the total number of line segments minus the number of sides.

Number of diagonals = (Total number of line segments connecting pairs of vertices) - (Number of sides)

Number of diagonals $= \binom{n}{2} - n$.

We are given that the number of diagonals is 44.

So, we have the equation:

Substitute the formula for $\binom{n}{2}$:

Multiply the entire equation by 2 to clear the denominator:

$n(n-1) - 2n = 88$

$n^2 - n - 2n = 88$

$n^2 - 3n = 88$

Rearrange the equation into a quadratic form:

We need to solve this quadratic equation for $n$. We can factor the quadratic expression. We look for two numbers that multiply to -88 and add to -3.

The factors of 88 are (1, 88), (2, 44), (4, 22), (8, 11).

We need one positive and one negative factor. The difference between the absolute values should be 3.

The pair (8, 11) works, as $8 - 11 = -3$ and $8 \times (-11) = -88$.

So, the quadratic equation can be factored as:

$(n - 11)(n + 8) = 0$

The possible values for $n$ are obtained by setting each factor to zero:

$n - 11 = 0 \implies n = 11$

$n + 8 = 0 \implies n = -8$

The number of sides of a polygon must be a positive integer and at least 3 (for a convex polygon). Therefore, $n = -8$ is not a valid solution.

The number of sides of the polygon is $n = 11$.

We can verify the answer:

Number of diagonals of an 11-sided polygon $= \binom{11}{2} - 11$

$\binom{11}{2} = \frac{11 \times 10}{2 \times 1} = \frac{110}{2} = 55$.

Number of diagonals $= 55 - 11 = 44$.

This matches the given number of diagonals.

The number of sides of the convex polygon is 11.

Given:

Total number of mice = 18.

The mice are to be placed into three groups: two experimental groups and one control group.

All groups are equally large.

To Find:

The number of ways to place the mice into these three groups.

Solution:

Since there are 18 mice and they are divided into 3 equally large groups, the number of mice in each group is $18 / 3 = 6$.

So, we need to form three groups, each containing 6 mice.

The groups are distinguishable: two are labeled "experimental" and one is labeled "control". The two experimental groups are distinct from each other in terms of their label (Experimental Group 1 and Experimental Group 2, although the question just says "two experimental groups", implying they might be functionally similar but are distinct sets of mice). Let's assume the two experimental groups are distinguishable (e.g., one is used for Drug A and the other for Drug B).

Step 1: Select the mice for the first experimental group.

We need to choose 6 mice from the 18 available mice for the first experimental group.

The number of ways to do this is $\binom{18}{6}$.

$\binom{18}{6} = \frac{18!}{6!(18-6)!} = \frac{18!}{6!12!}$.

Step 2: Select the mice for the second experimental group.

After selecting 6 mice for the first experimental group, there are $18 - 6 = 12$ mice remaining.

We need to choose 6 mice from these 12 remaining mice for the second experimental group.

The number of ways to do this is $\binom{12}{6}$.

$\binom{12}{6} = \frac{12!}{6!(12-6)!} = \frac{12!}{6!6!}$.

Step 3: Select the mice for the control group.

After selecting 6 mice for the first experimental group and 6 for the second, there are $12 - 6 = 6$ mice remaining.

These 6 remaining mice automatically form the control group. We need to choose 6 mice from these 6.

The number of ways to do this is $\binom{6}{6}$.

$\binom{6}{6} = \frac{6!}{6!(6-6)!} = \frac{6!}{6!0!} = 1$.

The total number of ways to place the mice into three distinct groups (Experimental Group 1, Experimental Group 2, Control Group) of 6 is the product of the number of ways for each step.

Total ways $= \binom{18}{6} \times \binom{12}{6} \times \binom{6}{6}$

Total ways $= \frac{18!}{6!12!} \times \frac{12!}{6!6!} \times 1$

Notice that the $12!$ in the numerator of the first term cancels with the $12!$ in the denominator of the second term.

Total ways $= \frac{18!}{6! \times 6! \times 6!}$.

Now, let's consider if the two experimental groups are indistinguishable in terms of their label, but the mice within them are distinct, and the groups themselves are distinct sets of mice. The problem states "two experimental groups", implying they are distinct. If the two experimental groups were considered indistinguishable (e.g., forming just two sets of experimental mice), we would divide by $2!$ for those two groups. However, the standard interpretation of such problems in an experimental context is that the groups are distinct (e.g., assigned to different treatments or conditions).

The calculation $\frac{18!}{6!6!6!}$ is the number of ways to partition 18 distinct items into three distinct groups of 6. Since the groups are labeled (Experimental 1, Experimental 2, Control), this corresponds to the correct interpretation.

Calculate the value:

$\frac{18!}{6!6!6!} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12!}{720 \times 720 \times 12!} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{720 \times 720}$.

$6! = 720$.

$\frac{18!}{6!6!6!} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6! \times 6! \times 6!} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$.

Let's calculate $\binom{18}{6}$ and $\binom{12}{6}$:

$\binom{18}{6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{18}{6 \times 3} \times \frac{16}{4 \times 2} \times \frac{15}{5} \times 17 \times 14 \times 13 = 1 \times 2 \times 3 \times 17 \times 14 \times 13 = 6 \times 17 \times 14 \times 13 = 102 \times 182 = 18564$.

$\binom{12}{6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{12}{6 \times 2} \times \frac{10}{5} \times \frac{9}{3} \times \frac{8}{4} \times 11 \times 7 = 1 \times 2 \times 3 \times 2 \times 11 \times 7 = 12 \times 77 = 924$.

Total ways $= 18564 \times 924 \times 1$.

$18564 \times 924 = 17153136$.

The number of ways to place the mice into three distinguishable groups of 6 is 17,153,136.

If the two experimental groups were indistinguishable (which is less likely in an experimental setup context but possible interpretation), the number of ways would be $\frac{1}{2!} \times \frac{18!}{6!6!6!} = \frac{17153136}{2} = 8576568$. However, given the phrasing, treating the groups as distinct is the standard approach.

Given:

A bag contains:

• White marbles: 6
• Red marbles: 5

Total number of marbles $= 6 + 5 = 11$.

We need to draw 4 marbles from the bag.

To Find:

The number of ways to draw 4 marbles under different conditions.

Solution:

We will use the combination formula $\binom{n}{k}$ to find the number of ways to choose $k$ items from $n$ distinct items.

(a) The marbles can be of any colour.

In this case, there is no restriction on the colours of the 4 marbles drawn. We just need to choose 4 marbles from the total of 11 marbles.

Number of ways to choose 4 marbles from 11 $= \binom{11}{4}$.

$\binom{11}{4} = \frac{11!}{4!(11-4)!} = \frac{11!}{4!7!} = \frac{11 \times 10 \times 9 \times 8 \times 7!}{4 \times 3 \times 2 \times 1 \times 7!} = \frac{11 \times 10 \times 9 \times 8}{24}$.

$\frac{11 \times 10 \times 9 \times 8}{24} = \frac{11 \times 10 \times 3 \times 3 \times 2 \times 4}{3 \times 2 \times 4 \times 1} = 11 \times 10 \times 3 / 3 = 11 \times 10 \times 3 = 330$.

The number of ways to draw 4 marbles of any colour is 330.

(b) Two must be white and two red.

We need to choose exactly 2 white marbles from the 6 white marbles AND exactly 2 red marbles from the 5 red marbles.

Number of ways to choose 2 white marbles from 6 $= \binom{6}{2}$.

$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{30}{2} = 15$.

Number of ways to choose 2 red marbles from 5 $= \binom{5}{2}$.

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10$.

Since these selections are independent, the total number of ways to choose 2 white and 2 red marbles is the product of the number of ways for each selection.

Number of ways $= \binom{6}{2} \times \binom{5}{2} = 15 \times 10 = 150$.

The number of ways to draw two white and two red marbles is 150.

(c) They must all be of the same colour.

This means the 4 marbles drawn must be either all white OR all red.

Case 1: All 4 marbles are white.

We need to choose 4 white marbles from the 6 white marbles. This is possible since we have at least 4 white marbles.

Number of ways to choose 4 white marbles from 6 $= \binom{6}{4}$.

$\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5 \times 4!}{4! \times 2 \times 1} = \frac{30}{2} = 15$.

Case 2: All 4 marbles are red.

We need to choose 4 red marbles from the 5 red marbles. This is possible since we have at least 4 red marbles.

Number of ways to choose 4 red marbles from 5 $= \binom{5}{4}$.

$\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = \frac{5 \times 4!}{4! \times 1} = 5$.

The events in Case 1 and Case 2 are mutually exclusive (the drawn marbles are either all white or all red, but not both). Therefore, the total number of ways is the sum of the ways in each case.

Number of ways $= (\text{Ways to draw 4 white}) + (\text{Ways to draw 4 red})$

Number of ways $= \binom{6}{4} + \binom{5}{4} = 15 + 5 = 20$.

The number of ways to draw 4 marbles of the same colour is 20.

Given:

Total number of players = 16.

We need to select a football team of 11 players.

To Find:

The number of ways to select the team under different conditions.

Solution:

The number of ways to select 11 players from 16 without any restriction is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Number of ways to select 11 players from 16 $= \binom{16}{11}$.

$\binom{16}{11} = \binom{16}{16-11} = \binom{16}{5}$.

$\binom{16}{5} = \frac{16!}{5!(16-5)!} = \frac{16!}{5!11!} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11!}{5 \times 4 \times 3 \times 2 \times 1 \times 11!} = \frac{16 \times 15 \times 14 \times 13 \times 12}{120}$.

$\frac{16 \times 15 \times 14 \times 13 \times 12}{120} = \frac{16 \times (3 \times 5) \times 14 \times 13 \times (3 \times 4)}{5 \times 4 \times 3 \times 2 \times 1} = \frac{16 \times 3 \times 14 \times 13 \times 12}{24}$.

$\frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1} = \frac{16 \times 15}{5 \times 3} \times \frac{12}{4 \times 2} \times 14 \times 13 = 16 \times 1 \times \frac{12}{8} \times 14 \times 13$. This simplification is incorrect.

Let's simplify properly:

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1}$

Cancel terms: $5 \times 3 = 15$ (cancels with 15 in numerator), $4 \times 2 = 8$ (cancels with 16 to leave 2).

$\binom{16}{5} = \frac{16}{4 \times 2} \times \frac{15}{5 \times 3} \times 14 \times 13 \times 12 = 2 \times 1 \times 14 \times 13 \times 12$. This is also incorrect.

Let's simplify again:

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1}$

$5 \times 2 = 10$, $4 \times 3 = 12$. Denominator $= 10 \times 12 = 120$.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{120}$.

$\frac{12}{120} = \frac{1}{10}$.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13}{10}$.

$\frac{16}{2} \times \frac{15}{5} \times 14 \times 13 = 8 \times 3 \times 14 \times 13$. Wait, the denominator is 10, not $5 \times 2$. Let's do it step by step.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{120}$.

$15 \times 8 = 120$. So, $\frac{15 \times 12}{120} = \frac{180}{120} = \frac{18}{12} = \frac{3}{2}$. Incorrect.

$\frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1}$.

$5 \times 4 \times 3 \times 2 \times 1 = 120$.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{120}$.

$\frac{12}{12} = 1$. So, $\frac{12}{120} = \frac{1}{10}$.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13}{10}$.

$\frac{10}{5} = 2$. So $\frac{15}{5} = 3$ and $\frac{16}{2} = 8$. Let's rewrite the denominator as $5 \times 2 \times 4 \times 3 \times 1$.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1} = \frac{16}{(4 \times 2)} \times \frac{15}{(5 \times 3)} \times 14 \times 13 \times 12 = 2 \times 1 \times 14 \times 13 \times 12$. Still wrong.

$\binom{16}{5} = \frac{16}{4} \times \frac{15}{5 \times 3} \times \frac{12}{2} \times 14 \times 13 = 4 \times 1 \times 6 \times 14 \times 13$. Incorrect.

Let's simplify the denominator: $5 \times 4 \times 3 \times 2 \times 1 = 120$.

Numerator: $16 \times 15 \times 14 \times 13 \times 12 = 5765760$.

$\frac{5765760}{120} = 48048$.

Let's try canceling from the fraction $\frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1}$.

$\frac{15}{5 \times 3} = 1$. $\frac{12}{4 \times 1} = 3$. $\frac{16}{2} = 8$.

So, $\binom{16}{5} = 8 \times 1 \times 14 \times 13 \times 3$. Still wrong.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{120}$.

$120 = 12 \times 10$. $\frac{12}{120} = \frac{1}{10}$.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13}{10}$.

$16 \times 15 = 240$. $\frac{240}{10} = 24$.

$\binom{16}{5} = 24 \times 14 \times 13$. This is also incorrect.

Let's use the prime factors of the denominator $5 \times 4 \times 3 \times 2 \times 1 = 5 \times (2 \times 2) \times 3 \times 2 \times 1 = 2^3 \times 3 \times 5$.

Numerator: $16 \times 15 \times 14 \times 13 \times 12 = (2^4) \times (3 \times 5) \times (2 \times 7) \times 13 \times (2^2 \times 3) = 2^{4+1+2} \times 3^{1+1} \times 5^1 \times 7^1 \times 13^1 = 2^7 \times 3^2 \times 5 \times 7 \times 13$.

$\binom{16}{5} = \frac{2^7 \times 3^2 \times 5 \times 7 \times 13}{2^3 \times 3 \times 5} = 2^{7-3} \times 3^{2-1} \times 5^{1-1} \times 7 \times 13 = 2^4 \times 3^1 \times 5^0 \times 7 \times 13 = 16 \times 3 \times 1 \times 7 \times 13$.

$16 \times 3 = 48$. $7 \times 13 = 91$.

$\binom{16}{5} = 48 \times 91$.

$48 \times 91 = 48 \times (90 + 1) = 48 \times 90 + 48 \times 1 = 4320 + 48 = 4368$.

The total number of ways to select 11 players from 16 is 4368.

(i) include 2 particular players?

Let the 2 particular players be P1 and P2. These 2 players must be included in the team of 11.

So, we have already selected 2 players (P1 and P2) for the team.

We need to select the remaining $11 - 2 = 9$ players for the team.

These 9 players must be chosen from the remaining $16 - 2 = 14$ players (excluding P1 and P2).

The number of ways to choose 9 players from 14 is given by $\binom{14}{9}$.

$\binom{14}{9} = \binom{14}{14-9} = \binom{14}{5}$.

$\binom{14}{5} = \frac{14!}{5!(14-5)!} = \frac{14!}{5!9!} = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9!}{5 \times 4 \times 3 \times 2 \times 1 \times 9!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{120}$.

$\frac{14 \times 13 \times 12 \times 11 \times 10}{120} = \frac{12}{120} \times 14 \times 13 \times 11 \times 10 = \frac{1}{10} \times 14 \times 13 \times 11 \times 10 = 14 \times 13 \times 11$.

$14 \times 13 = 182$.

$182 \times 11 = 182 \times (10 + 1) = 1820 + 182 = 2002$.

The number of ways to select the team including 2 particular players is 2002.

(ii) exclude 2 particular players?

Let the 2 particular players be P1 and P2. These 2 players must be excluded from the team of 11.

This means the team must be selected from the remaining $16 - 2 = 14$ players (excluding P1 and P2).

We need to select all 11 players for the team from these 14 players.

The number of ways to choose 11 players from 14 is given by $\binom{14}{11}$.

$\binom{14}{11} = \binom{14}{14-11} = \binom{14}{3}$.

$\binom{14}{3} = \frac{14!}{3!(14-3)!} = \frac{14!}{3!11!} = \frac{14 \times 13 \times 12 \times 11!}{3 \times 2 \times 1 \times 11!} = \frac{14 \times 13 \times 12}{6}$.

$\frac{14 \times 13 \times 12}{6} = 14 \times 13 \times \frac{12}{6} = 14 \times 13 \times 2$.

$14 \times 13 = 182$.

$182 \times 2 = 364$.

The number of ways to select the team excluding 2 particular players is 364.

Given:

Number of students in Class XI = 20.

Number of students in Class XII = 20.

We need to constitute a sports team of 11 students.

Constraint: The team must include at least 5 students from Class XI and at least 5 students from Class XII.

To Find:

The number of ways to constitute the team under the given constraints.

Solution:

Let $n_{XI}$ be the number of students chosen from Class XI and $n_{XII}$ be the number of students chosen from Class XII.

The total number of students in the team is 11, so we must have $n_{XI} + n_{XII} = 11$.

The constraints are:

• $n_{XI} \ge 5$
• $n_{XII} \ge 5$

Since the total number of students in each class is 20, we also have $n_{XI} \le 20$ and $n_{XII} \le 20$. However, given $n_{XI} + n_{XII} = 11$, these upper bounds (20) are automatically satisfied because the maximum value $n_{XI}$ can take is 6 (when $n_{XII}=5$) and the maximum value $n_{XII}$ can take is 6 (when $n_{XI}=5$), both of which are less than or equal to 20.

We need to find pairs of integers $(n_{XI}, n_{XII})$ such that:

$n_{XI} + n_{XII} = 11$

$n_{XI} \ge 5$

$n_{XII} \ge 5$

Let's list the possible integer pairs $(n_{XI}, n_{XII})$ that sum to 11, starting with $n_{XI} = 5$ (the minimum allowed value):

• If $n_{XI} = 5$, then $n_{XII} = 11 - 5 = 6$. This satisfies $n_{XII} \ge 5$. So, $(5, 6)$ is a valid combination.
• If $n_{XI} = 6$, then $n_{XII} = 11 - 6 = 5$. This satisfies $n_{XII} \ge 5$. So, $(6, 5)$ is a valid combination.

If we consider $n_{XI} \ge 7$, then $n_{XII} = 11 - n_{XI} \le 11 - 7 = 4$, which violates $n_{XII} \ge 5$. Similarly, if we consider $n_{XII} \ge 7$, then $n_{XI} = 11 - n_{XII} \le 11 - 7 = 4$, which violates $n_{XI} \ge 5$.

Thus, the only possible combinations for the number of students from each class are (5 from Class XI, 6 from Class XII) and (6 from Class XI, 5 from Class XII).

We calculate the number of ways for each allowed combination:

Case 1: Choosing 5 students from Class XI and 6 students from Class XII ($n_{XI}=5, n_{XII}=6$).

Number of ways to choose 5 students from 20 (Class XI) $= \binom{20}{5}$.

Number of ways to choose 6 students from 20 (Class XII) $= \binom{20}{6}$.

Number of ways for Case 1 $= \binom{20}{5} \times \binom{20}{6}$.

$\binom{20}{5} = \frac{20!}{5!15!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = \frac{20}{5 \times 4} \times \frac{18}{3 \times 2} \times 19 \times 17 \times 16 = 1 \times 3 \times 19 \times 17 \times 16 = 3 \times 19 \times 17 \times 16 = 3 \times 323 \times 16 = 969 \times 16 = 15504$.

$\binom{20}{6} = \frac{20!}{6!14!} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{20}{5 \times 4} \times \frac{18}{6 \times 3} \times \frac{16}{2} \times 19 \times 17 \times 15 = 1 \times 1 \times 8 \times 19 \times 17 \times 15 = 8 \times 323 \times 15 = 8 \times 4845 = 38760$.

Number of ways for Case 1 $= 15504 \times 38760$.

$15504 \times 38760 = 601807440$. This number seems very large. Let's recheck the calculations.

Let's calculate $\binom{20}{5}$ again:

$\binom{20}{5} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = \frac{20}{5 \times 4} \times \frac{18}{3 \times 2} \times 19 \times 17 \times 16 = 1 \times 3 \times 19 \times 17 \times 16 = 3 \times 19 \times 17 \times 16 = 57 \times 272 = 15504$. This is correct.

Let's calculate $\binom{20}{6}$ again:

$\binom{20}{6} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{20}{5 \times 4} \times \frac{18}{6 \times 3 \times 1} \times \frac{16}{2} \times 19 \times 17 \times 15 = 1 \times 1 \times 8 \times 19 \times 17 \times 15 = 8 \times 19 \times 17 \times 15 = 8 \times 323 \times 15 = 8 \times 4845 = 38760$. This is correct.

Number of ways for Case 1 $= 15504 \times 38760 = 601807440$. This still seems large for typical combinatorics problems at this level, but the calculation steps are correct. Let me double-check the problem statement and constraints. Everything seems correct. It's possible the number is indeed this large.

Case 2: Choosing 6 students from Class XI and 5 students from Class XII ($n_{XI}=6, n_{XII}=5$).

Number of ways to choose 6 students from 20 (Class XI) $= \binom{20}{6} = 38760$.

Number of ways to choose 5 students from 20 (Class XII) $= \binom{20}{5} = 15504$.

Number of ways for Case 2 $= \binom{20}{6} \times \binom{20}{5} = 38760 \times 15504 = 601807440$.

The total number of ways to constitute the team is the sum of the ways for Case 1 and Case 2.

Total number of ways = Ways (Case 1) + Ways (Case 2)

Total number of ways $= 601807440 + 601807440 = 2 \times 601807440 = 1203614880$.

The number of ways the team can be constituted is 1,203,614,880.

Given:

Number of girls = 4.

Number of boys = 7.

Total number of members in the group $= 4 + 7 = 11$.

We need to select a team of 5 members.

To Find:

The number of ways to select a team of 5 members under different conditions.

Solution:

We will use the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ to find the number of ways to choose $k$ items from $n$ distinct items.

(i) The team has no girls.

If the team has no girls, all 5 members must be boys. We need to select 5 boys from the 7 available boys.

Number of ways to choose 5 boys from 7 $= \binom{7}{5}$.

Since $\binom{n}{k} = \binom{n}{n-k}$, $\binom{7}{5} = \binom{7}{7-5} = \binom{7}{2}$.

$\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6 \times 5!}{2 \times 1 \times 5!} = \frac{7 \times 6}{2} = 21$.

The number of ways to select a team with no girls is 21.

(ii) The team has at least one boy and one girl.

We need to select a team of 5 members such that there is at least 1 boy and at least 1 girl.

The possible combinations of (Girls, Boys) that sum to 5, satisfying the conditions (number of girls $\ge 1$, number of boys $\ge 1$, number of girls $\le 4$, number of boys $\le 7$) are:

• 1 girl and 4 boys
• 2 girls and 3 boys
• 3 girls and 2 boys
• 4 girls and 1 boy

Calculate the number of ways for each combination:

• 1 girl and 4 boys: $\binom{4}{1} \times \binom{7}{4} = 4 \times 35 = 140$. ($\binom{7}{4} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$)
• 2 girls and 3 boys: $\binom{4}{2} \times \binom{7}{3} = 6 \times 35 = 210$. ($\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$)
• 3 girls and 2 boys: $\binom{4}{3} \times \binom{7}{2} = 4 \times 21 = 84$. ($\binom{4}{3} = 4$, $\binom{7}{2} = 21$)
• 4 girls and 1 boy: $\binom{4}{4} \times \binom{7}{1} = 1 \times 7 = 7$. ($\binom{4}{4} = 1$, $\binom{7}{1} = 7$)

The total number of ways for this case is the sum of the ways for these combinations:

Total ways $= 140 + 210 + 84 + 7 = 441$.

Alternate approach for (ii):

Find the total number of ways to select 5 members from the 11 without any restriction, and subtract the cases where the condition is not met (i.e., cases where there are no girls or no boys).

Total number of ways to select 5 members from 11 $= \binom{11}{5}$.

$\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = \frac{11 \times (5 \times 2) \times (3 \times 3) \times (4 \times 2) \times 7}{5 \times 4 \times 3 \times 2 \times 1}$.

$\binom{11}{5} = \frac{11 \times \cancel{10}^2 \times \cancel{9}^3 \times \cancel{8}^2 \times 7}{\cancel{5} \times \cancel{4} \times \cancel{3} \times \cancel{2} \times 1} = 11 \times 2 \times 3 \times 2 \times 7$. Incorrect cancellation.

$\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{120}$.

$\frac{10 \times 9 \times 8 \times 7}{120} = \frac{5040}{120} = 42$.

$\binom{11}{5} = 11 \times 42 = 462$.

Total ways to select 5 from 11 is 462.

Cases that violate "at least one boy and one girl":

- No girls (all boys): We need to choose 5 boys from 7 and 0 girls from 4. Ways $= \binom{7}{5} \times \binom{4}{0} = 21 \times 1 = 21$.

- No boys (all girls): We need to choose 5 girls from 4 and 0 boys from 7. This is impossible since there are only 4 girls. Ways $= \binom{4}{5} \times \binom{7}{0} = 0 \times 1 = 0$.

Number of ways with at least one boy and one girl = (Total ways) - (Ways with no girls) - (Ways with no boys)

Number of ways $= 462 - 21 - 0 = 441$.

Both approaches give the same result. The number of ways to select a team with at least one boy and one girl is 441.

(iii) The team has at least three girls.

We need to select a team of 5 members such that there are at least 3 girls.

The possible combinations of (Girls, Boys) that sum to 5, satisfying the condition (number of girls $\ge 3$) and the constraints (number of girls $\le 4$, number of boys $\le 7$) are:

• 3 girls and 2 boys
• 4 girls and 1 boy

Calculate the number of ways for each combination:

• 3 girls and 2 boys: $\binom{4}{3} \times \binom{7}{2} = 4 \times 21 = 84$.
• 4 girls and 1 boy: $\binom{4}{4} \times \binom{7}{1} = 1 \times 7 = 7$.

The total number of ways for this case is the sum of the ways for these combinations:

Total ways $= 84 + 7 = 91$.

The number of ways to select a team with at least three girls is 91.

Given:

To Find:

The value of $n$.

Solution:

We use the property of combinations which states that if $\binom{n}{a} = \binom{n}{b}$ and $a \neq b$, then $a+b = n$.

In the given equation, we have $a = 12$ and $b = 8$.

Since $12 \neq 8$, we must have $12 + 8 = n$.

$n = 12 + 8$

$n = 20$

The value of $n$ is 20.

Comparing this with the given options:

(A) 20

(B) 12

(C) 6

(D) 30

The correct option is (A).

The final answer is $\boxed{20}$.

Given:

A coin is tossed 6 times.

To Find:

The number of possible outcomes.

Solution:

When a coin is tossed, there are two possible outcomes: Heads (H) or Tails (T).

Since the coin is tossed 6 times independently, the total number of possible outcomes is the product of the number of outcomes for each toss.

For the first toss, there are 2 outcomes.

For the second toss, there are 2 outcomes.

... and so on, for all 6 tosses.

The total number of possible outcomes = $2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.

Calculate the value of $2^6$:

$2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 4 \times 4 \times 4 = 16 \times 4 = 64$.

The number of possible outcomes when a coin is tossed 6 times is 64.

Comparing this with the given options:

(A) 36

(B) 64

(C) 12

(D) 32

The correct option is (B).

The final answer is $\boxed{64}$.

Given:

Set of distinct digits = {2, 3, 4, 7}.

We need to form four-digit numbers.

Constraint: Each digit must be used only once.

To Find:

The number of different four-digit numbers that can be formed.

Solution:

We have a set of 4 distinct digits {2, 3, 4, 7}.

We need to arrange these 4 digits to form a four-digit number. Since each digit must be used only once, this is a permutation problem where we are arranging all 4 distinct digits.

The number of permutations of $n$ distinct objects is $n!$.

In this case, $n = 4$ (the number of distinct digits).

The number of different four-digit numbers that can be formed is $4!$.

Calculate $4!$:

$4! = 4 \times 3 \times 2 \times 1 = 24$.

The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is 24.

Comparing this with the given options:

(A) 120

(B) 96

(C) 24

(D) 100

The correct option is (C).

The final answer is $\boxed{24}$.

Given:

Set of distinct digits = {3, 4, 5, 6}.

We are forming numbers using all these 4 digits at a time without repetition.

To Find:

The sum of the digits in the unit place of all the numbers formed.

Solution:

The numbers formed are 4-digit numbers using the digits 3, 4, 5, and 6 exactly once.

The total number of such 4-digit numbers is the number of permutations of 4 distinct digits, which is $4! = 24$.

These 24 numbers are formed by arranging the digits 3, 4, 5, and 6 in all possible ways.

We are interested in the sum of the digits that appear in the unit place of these 24 numbers.

Let's consider the positions in the 4-digit number: Thousands, Hundreds, Tens, Units.

Any of the four digits {3, 4, 5, 6} can appear in the unit place.

Let's count how many times each digit appears in the unit place.

If a specific digit (say 3) is in the unit place, the remaining 3 digits ({4, 5, 6}) can be arranged in the remaining 3 positions (Thousands, Hundreds, Tens) in $3!$ ways.

Number of ways to arrange the remaining 3 digits $= 3! = 3 \times 2 \times 1 = 6$.

So, the digit 3 appears in the unit place 6 times.

Similarly, the digit 4 appears in the unit place 6 times.

The digit 5 appears in the unit place 6 times.

The digit 6 appears in the unit place 6 times.

The sum of the digits in the unit place is the sum of the values of the digits multiplied by the number of times they appear in the unit place.

Sum of digits in unit place = (3 $\times$ number of times 3 is in unit place) + (4 $\times$ number of times 4 is in unit place) + (5 $\times$ number of times 5 is in unit place) + (6 $\times$ number of times 6 is in unit place)

Sum of digits in unit place $= (3 \times 6) + (4 \times 6) + (5 \times 6) + (6 \times 6)$

Sum of digits in unit place $= 18 + 24 + 30 + 36$.

Sum of digits in unit place $= 6 \times (3 + 4 + 5 + 6)$

Sum of digits in unit place $= 6 \times (18)$

$6 \times 18 = 108$.

The sum of the digits in the unit place of all the numbers formed is 108.

Comparing this with the given options:

(A) 432

(B) 108

(C) 36

(D) 18

The correct option is (B).

The final answer is $\boxed{108}$.

Given:

Number of vowels available = 4.

Number of consonants available = 5.

We need to form words using 2 vowels and 3 consonants.

To Find:

The total number of words formed by selecting 2 vowels from 4 and 3 consonants from 5.

Solution:

The process of forming a word involves two steps: selecting the letters and then arranging them.

Step 1: Select the vowels and consonants.

We need to choose 2 vowels from the 4 available vowels. The number of ways to do this is $\binom{4}{2}$.

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} = \frac{12}{2} = 6$.

We need to choose 3 consonants from the 5 available consonants. The number of ways to do this is $\binom{5}{3}$.

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{20}{2} = 10$.

The total number of ways to select 2 vowels and 3 consonants is the product of the number of ways to make each selection.

Number of ways to select letters $= \binom{4}{2} \times \binom{5}{3} = 6 \times 10 = 60$.

After this step, we have a set of $2 + 3 = 5$ distinct letters (the chosen vowels and consonants).

Step 2: Arrange the selected letters to form words.

We have a set of 5 distinct letters. The number of different words that can be formed by arranging these 5 distinct letters is the number of permutations of 5 distinct items, which is $5!$.

Number of ways to arrange the 5 selected letters $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

The total number of different words that can be formed is the product of the number of ways to select the letters and the number of ways to arrange the selected letters.

Total number of words = (Number of ways to select letters) $\times$ (Number of ways to arrange letters)

Total number of words $= (\binom{4}{2} \times \binom{5}{3}) \times 5! = 60 \times 120$.

$60 \times 120 = 7200$.

The total number of words formed is 7200.

Comparing this with the given options:

(A) 60

(B) 120

(C) 7200

(D) 720

The correct option is (C).

The final answer is $\boxed{7200}$.

Given:

Set of digits = {0, 1, 2, 3, 4, 5}. There are 6 distinct digits.

We need to form a five-digit number.

The number must be divisible by 3.

No digit should be repeated.

To Find:

The total number of ways this can be done.

Solution:

A number is divisible by 3 if the sum of its digits is divisible by 3.

We need to form a five-digit number using 5 distinct digits chosen from the set {0, 1, 2, 3, 4, 5}. The sum of the digits of the five-digit number must be divisible by 3.

The sum of all available digits = $0 + 1 + 2 + 3 + 4 + 5 = 15$.

We are forming a 5-digit number, which means we are choosing 5 digits from the 6 available digits. The digit that is NOT included in the 5-digit number determines the sum of the digits in the number.

Let $S$ be the sum of the digits of the five-digit number. $S = 15 - (\text{the digit excluded})$.

For the five-digit number to be divisible by 3, the sum of its digits, $S$, must be divisible by 3.

Since 15 is divisible by 3, $S = 15 - (\text{the digit excluded})$ will be divisible by 3 if and only if the excluded digit is divisible by 3.

The digits in the set {0, 1, 2, 3, 4, 5} that are divisible by 3 are 0 and 3.

So, for the sum of the digits of the 5-digit number to be divisible by 3, we must exclude either the digit 0 or the digit 3 from the original set.

Case 1: The digit 0 is excluded.

The set of 5 digits used for the number is {1, 2, 3, 4, 5}. The sum of these digits is $1 + 2 + 3 + 4 + 5 = 15$, which is divisible by 3.

We need to form a five-digit number using these 5 distinct digits. Since none of these digits is 0, any arrangement of these 5 digits will form a 5-digit number.

The number of 5-digit numbers formed by arranging 5 distinct digits is $5!$.

Number of ways for Case 1 $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Case 2: The digit 3 is excluded.

The set of 5 digits used for the number is {0, 1, 2, 4, 5}. The sum of these digits is $0 + 1 + 2 + 4 + 5 = 12$, which is divisible by 3.

We need to form a five-digit number using these 5 distinct digits. However, one of the digits is 0, and the first digit of a five-digit number cannot be 0.

We have 5 positions for the digits $d_1 d_2 d_3 d_4 d_5$.

The first digit $d_1$ can be any of the 4 non-zero digits from the set {1, 2, 4, 5}. So there are 4 choices for $d_1$.

The remaining 4 positions ($d_2, d_3, d_4, d_5$) must be filled with the remaining 4 digits from the set {0, 1, 2, 4, 5} (excluding the digit used for $d_1$). These 4 remaining digits include 0.

The number of ways to arrange the remaining 4 distinct digits in the remaining 4 positions is $4!$.

Number of ways to fill the remaining 4 positions $= 4! = 4 \times 3 \times 2 \times 1 = 24$.

The number of 5-digit numbers formed in Case 2 is the number of choices for the first digit multiplied by the number of ways to arrange the remaining digits.

Number of ways for Case 2 $= (\text{Choices for } d_1) \times (\text{Ways to arrange remaining 4 digits}) = 4 \times 4! = 4 \times 24 = 96$.

Another way to calculate Case 2: Total permutations of {0, 1, 2, 4, 5} is $5! = 120$. The number of permutations starting with 0 is $4! = 24$. The number of 5-digit numbers is $5! - 4! = 120 - 24 = 96$.

The total number of ways to form a five-digit number divisible by 3 without repetition is the sum of the ways in Case 1 and Case 2.

Total number of ways = Ways (Case 1) + Ways (Case 2)

Total number of ways $= 120 + 96 = 216$.

The total number of ways this can be done is 216.

Comparing this with the given options:

(A) 216

(B) 600

(C) 240

(D) 3125

The correct option is (A).

The final answer is $\boxed{216}$.

Given:

Everybody in a room shakes hands with everybody else exactly once.

The total number of handshakes is 66.

To Find:

The total number of persons in the room.

Solution:

Let the total number of persons in the room be $n$.

A handshake occurs between two distinct persons. Since everybody shakes hands with everybody else, every pair of distinct persons will result in exactly one handshake.

The total number of handshakes is the number of ways to choose 2 persons from the $n$ persons, without regard to the order (since shaking hand between person A and person B is the same handshake as between person B and person A).

This is a combination problem. The number of ways to choose 2 persons from $n$ is given by $\binom{n}{2}$.

$\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)(n-2)!}{2 \times 1 \times (n-2)!} = \frac{n(n-1)}{2}$.

We are given that the total number of handshakes is 66.

So, we have the equation:

Multiply both sides by 2:

$n(n-1) = 66 \times 2$

$n(n-1) = 132$

This is a quadratic equation in $n$. Expand and rearrange:

$n^2 - n = 132$

We need to solve this quadratic equation for $n$. We are looking for two consecutive integers whose product is 132. We can factor 132.

$132 = 1 \times 132$

$132 = 2 \times 66$

$132 = 3 \times 44$

$132 = 4 \times 33$

$132 = 6 \times 22$

$132 = 11 \times 12$

The consecutive integers are 11 and 12. Since $n(n-1) = 132$, and we assume $n$ is positive, $n$ must be the larger of the two consecutive integers. So, $n=12$ and $n-1=11$.

Let's solve the quadratic equation by factoring:

We need two numbers that multiply to -132 and add to -1. The numbers are 11 and -12.

$(n + 11)(n - 12) = 0$

The possible values for $n$ are:

$n + 11 = 0 \implies n = -11$

$n - 12 = 0 \implies n = 12$

Since the number of persons must be a positive integer, $n = 12$ is the valid solution.

The total number of persons in the room is 12.

Comparing this with the given options:

(A) 11

(B) 12

(C) 13

(D) 14

The correct option is (B).

The final answer is $\boxed{12}$.

Given:

Total number of points in a set = 12.

Number of collinear points among these 12 = 7.

The remaining $12 - 7 = 5$ points are non-collinear with respect to each other and also not on the line containing the 7 collinear points.

To Find:

The number of triangles that can be formed by choosing vertices from this set of 12 points.

Solution:

A triangle is formed by choosing any three non-collinear points from the given set of points.

The total number of ways to choose 3 points from the 12 points without any restriction is given by $\binom{12}{3}$.

$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} = \frac{12 \times 11 \times 10}{6}$.

$\frac{12 \times 11 \times 10}{6} = \frac{12}{6} \times 11 \times 10 = 2 \times 11 \times 10 = 220$.

This total includes combinations of 3 points that form a triangle AND combinations of 3 points that lie on the same line (collinear points, which do not form a triangle).

The 7 collinear points lie on the same line. If we choose any 3 points from these 7 collinear points, they will not form a triangle.

The number of ways to choose 3 points from the 7 collinear points is given by $\binom{7}{3}$.

$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} = \frac{7 \times 6 \times 5}{6} = 35$.

These 35 combinations of points do not form triangles.

The number of triangles is the total number of ways to choose 3 points minus the number of ways to choose 3 collinear points.

Number of triangles = (Total ways to choose 3 points) - (Ways to choose 3 collinear points)

Number of triangles $= \binom{12}{3} - \binom{7}{3} = 220 - 35 = 185$.

The number of triangles that are formed is 185.

Comparing this with the given options:

(A) 105

(B) 15

(C) 175

(D) 185

The correct option is (D).

The final answer is $\boxed{185}$.

Given:

Set 1: Four parallel lines.

Set 2: Three parallel lines.

The lines in Set 1 intersect the lines in Set 2.

To Find:

The number of parallelograms that can be formed by these intersecting lines.

Solution:

A parallelogram is formed by two pairs of parallel lines. In this scenario, the two pairs of parallel lines must come from the two given sets of parallel lines.

Specifically, a parallelogram will be formed by choosing two lines from the set of four parallel lines and two lines from the set of three parallel lines.

The number of ways to choose 2 lines from the set of 4 parallel lines is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Number of ways to choose 2 lines from 4 $= \binom{4}{2}$.

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} = \frac{12}{2} = 6$.

The number of ways to choose 2 lines from the set of 3 parallel lines is given by $\binom{3}{2}$.

$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2!}{2! \times 1} = 3$.

Each combination of 2 lines from the first set and 2 lines from the second set forms a unique parallelogram.

The total number of parallelograms is the product of the number of ways to choose the lines from each set.

Number of parallelograms = (Number of ways to choose 2 lines from 4) $\times$ (Number of ways to choose 2 lines from 3)

Number of parallelograms $= \binom{4}{2} \times \binom{3}{2} = 6 \times 3 = 18$.

The number of parallelograms that can be formed is 18.

Comparing this with the given options:

(A) 6

(B) 18

(C) 12

(D) 9

The correct option is (B).

The final answer is $\boxed{18}$.

Given:

Total number of players = 22.

We need to select a team of 11 players.

Constraint 1: 2 particular players must always be included in the team.

Constraint 2: 4 particular players must always be excluded from the team.

To Find:

The number of ways to select the team under the given constraints.

Solution:

We need to select a team of 11 players with specific conditions on certain players.

Let the 2 players who must be included be $I_1$ and $I_2$.

Let the 4 players who must be excluded be $E_1, E_2, E_3, E_4$.

Since $I_1$ and $I_2$ must always be included, we can consider them as already selected for the team. This accounts for 2 members of the team.

We need to select the remaining $11 - 2 = 9$ players for the team.

Since $E_1, E_2, E_3, E_4$ must always be excluded, they are not available for selection. These 4 players are removed from the original pool of 22 players.

The 2 players who must be included ($I_1, I_2$) are also not available for selection of the remaining players, as they are already selected.

So, the pool of available players from which we need to choose the remaining 9 players is the original 22 players minus the 2 included players and minus the 4 excluded players.

Number of available players for the remaining selection = $22 - 2 - 4 = 16$.

We need to choose 9 players from these 16 available players to complete the team of 11.

The number of ways to choose 9 players from 16 is given by the combination formula $\binom{16}{9}$.

$\binom{16}{9} = \frac{16!}{9!(16-9)!} = \frac{16!}{9!7!}$.

Using the property $\binom{n}{k} = \binom{n}{n-k}$, we have $\binom{16}{9} = \binom{16}{16-9} = \binom{16}{7}$.

$\binom{16}{7} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{5040}$.

Let's calculate the value or compare with options.

The number of ways is $\binom{16}{9}$. Let's check the options.

(A) $\binom{16}{11}$

(B) $\binom{16}{5}$

(C) $\binom{16}{9}$

(D) $\binom{20}{9}$

Our result is $\binom{16}{9}$, which matches option (C).

Let's quickly calculate $\binom{16}{9}$ and $\binom{16}{5}$ and $\binom{16}{11}$.

$\binom{16}{9} = \binom{16}{7} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$.

$7 \times 2 = 14$ (cancels with 14).

$6 \times 5 = 30$. $30 \times 4 \times 3 = 360$. $360 \times 1 = 360$. Denominator $= 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$.

$\frac{16 \times 15 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$ (after canceling 14 with $7 \times 2$).

$\frac{15}{5 \times 3} = 1$. $\frac{12}{6 \times 2} = 1$. $\frac{16}{4} = 4$.

$\binom{16}{7} = 4 \times 1 \times 1 \times 13 \times 11 \times 10 = 4 \times 130 \times 11 = 520 \times 11 = 5720$.

$\binom{16}{5} = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1} = 4368$ (from Question 23).

$\binom{16}{11} = \binom{16}{5} = 4368$.

The number of ways is $\binom{16}{9} = 5720$. Let's recheck the options against our calculated value. The options are given in combination notation.

Our result is $\binom{16}{9}$. This matches option (C).

The number of ways is $\binom{16}{9}$.

Comparing this with the given options:

(A) $\binom{16}{11}$

(B) $\binom{16}{5}$

(C) $\binom{16}{9}$

(D) $\binom{20}{9}$

The correct option is (C).

The final answer is $\boxed{^{16}C_9}$.

Given:

We are considering 5-digit telephone numbers.

A 5-digit number can use digits from {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

We need to find the number of 5-digit telephone numbers having at least one of their digits repeated.

To Find:

The number of 5-digit telephone numbers with at least one repeated digit.

Solution:

A 5-digit telephone number is formed by arranging 5 digits. The first digit of a telephone number can typically be any digit from 0 to 9, unless otherwise specified. However, usually in number formation problems, the first digit cannot be 0 for it to be considered a 5-digit number. For telephone numbers, the first digit can be 0 in many systems. Let's assume the number can start with 0, forming a 5-digit sequence.

The condition "at least one of their digits repeated" is the complement of the condition "all digits are distinct (no repetition)".

Total number of 5-digit telephone numbers = (Total number of 5-digit numbers without repetition) + (Total number of 5-digit numbers with at least one repetition).

Number of 5-digit numbers with at least one repetition = (Total number of 5-digit telephone numbers) - (Number of 5-digit telephone numbers with distinct digits).

Step 1: Calculate the total number of 5-digit telephone numbers.

There are 5 positions for the digits.

For each position, there are 10 choices for the digit (0 to 9), and repetition is allowed.

Number of choices for the first digit = 10.

Number of choices for the second digit = 10.

Number of choices for the third digit = 10.

Number of choices for the fourth digit = 10.

Number of choices for the fifth digit = 10.

Total number of 5-digit telephone numbers $= 10 \times 10 \times 10 \times 10 \times 10 = 10^5 = 100,000$.

Step 2: Calculate the number of 5-digit telephone numbers with distinct digits.

We need to choose 5 distinct digits from the 10 available digits and arrange them in the 5 positions.

Number of choices for the first digit = 10 (any digit from 0 to 9).

Number of choices for the second digit = 9 (any digit except the first one).

Number of choices for the third digit = 8 (any digit except the first two).

Number of choices for the fourth digit = 7 (any digit except the first three).

Number of choices for the fifth digit = 6 (any digit except the first four).

Number of 5-digit telephone numbers with distinct digits $= 10 \times 9 \times 8 \times 7 \times 6$.

This is the number of permutations of 10 items taken 5 at a time, $P(10, 5)$.

$P(10, 5) = \frac{10!}{(10-5)!} = \frac{10!}{5!} = 10 \times 9 \times 8 \times 7 \times 6$.

$10 \times 9 = 90$. $8 \times 7 = 56$. $56 \times 6 = 336$.

$90 \times 336 = 9 \times 3360 = 30240$.

Number of 5-digit telephone numbers with distinct digits $= 30,240$.

Step 3: Calculate the number of 5-digit telephone numbers with at least one repeated digit.

Number of numbers with repetition = (Total number of numbers) - (Number of numbers with distinct digits)

Number of numbers with repetition $= 100,000 - 30,240$.

$100,000 - 30,240 = 69,760$.

The number of 5-digit telephone numbers having at least one of their digits repeated is 69,760.

Comparing this with the given options:

(A) 90,000

(B) 10,000

(C) 30,240

(D) 69,760

The correct option is (D).

The final answer is $\boxed{69,760}$.

Given:

Number of men = 4.

Number of women = 6.

We need to choose a committee.

Constraint 1: The committee must include at least two men.

Constraint 2: The committee must include exactly twice as many women as men.

To Find:

The number of ways to choose the committee under the given constraints.

Solution:

Let the number of men in the committee be $m$ and the number of women be $w$.

From Constraint 2, we have $w = 2m$.

From Constraint 1, we have $m \ge 2$.

Also, the number of men selected cannot exceed the total number of men available ($m \le 4$), and the number of women selected cannot exceed the total number of women available ($w \le 6$).

We need to find integer values of $m$ that satisfy these conditions:

• $m \ge 2$
• $m \le 4$
• $w = 2m \le 6 \implies m \le 3$

Combining these conditions, the possible values for $m$ are 2 and 3.

We consider the possible combinations of (men, women) based on the value of $m$:

Case 1: Number of men $m = 2$.

If $m = 2$, then $w = 2m = 2 \times 2 = 4$.

This combination is (2 men, 4 women). Let's check if this is possible with the available members:

• Number of men to choose = 2. Available men = 4. This is possible ($\binom{4}{2}$).
• Number of women to choose = 4. Available women = 6. This is possible ($\binom{6}{4}$).

The number of ways to select 2 men from 4 is $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.

The number of ways to select 4 women from 6 is $\binom{6}{4} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$.

The number of ways to form the committee in Case 1 is $\binom{4}{2} \times \binom{6}{4} = 6 \times 15 = 90$.

Case 2: Number of men $m = 3$.

If $m = 3$, then $w = 2m = 2 \times 3 = 6$.

This combination is (3 men, 6 women). Let's check if this is possible with the available members:

• Number of men to choose = 3. Available men = 4. This is possible ($\binom{4}{3}$).
• Number of women to choose = 6. Available women = 6. This is possible ($\binom{6}{6}$).

The number of ways to select 3 men from 4 is $\binom{4}{3} = 4$.

The number of ways to select 6 women from 6 is $\binom{6}{6} = 1$.

The number of ways to form the committee in Case 2 is $\binom{4}{3} \times \binom{6}{6} = 4 \times 1 = 4$.

If we consider $m=4$, then $w=2m=8$. We only have 6 women available, so we cannot choose 8 women. Thus, $m=4$ is not possible.

The total number of ways to choose the committee is the sum of the ways in Case 1 and Case 2, as these cases are mutually exclusive.

Total number of ways = Ways (Case 1) + Ways (Case 2)

Total number of ways $= 90 + 4 = 94$.

The number of ways to choose the committee is 94.

Comparing this with the given options:

(A) 94

(B) 126

(C) 128

(D) None

The correct option is (A).

The final answer is $\boxed{94}$.

Given:

We need to form 9-digit numbers.

The digits must all be different.

The available digits are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. There are 10 distinct digits.

To Find:

The total number of 9-digit numbers which have all different digits.

Solution:

A 9-digit number is formed by arranging 9 distinct digits in 9 positions ($d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9$).

The digits must be chosen from the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Since the number has 9 digits and they are all different, we must choose 9 distinct digits from these 10 available digits.

We are forming a 9-digit number, which means the first digit cannot be 0.

We can solve this by considering the positions and the choices for each position, or by considering the set of digits used and their arrangement.

Method 1: By positions.

Position $d_1$ (the first digit): It cannot be 0. So there are 9 choices for the first digit (1 to 9).

Position $d_2$: We need to choose a digit different from the first digit. There are 9 remaining digits available (including 0). So there are 9 choices for the second digit.

Position $d_3$: We need to choose a digit different from the first two digits. There are 8 remaining digits available. So there are 8 choices for the third digit.

... and so on.

Position $d_9$: We need to choose a digit different from the first eight digits. There are $10 - 8 = 2$ remaining digits available. So there are 2 choices for the ninth digit.

The total number of such 9-digit numbers is the product of the number of choices for each position:

Total numbers $= 9 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2$.

This can be written as $9 \times (9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2)$.

The term in the parenthesis is $9!$.

Total numbers $= 9 \times 9!$.

Method 2: By selecting digits and arranging.

We need to choose 9 distinct digits from the 10 available digits {0, 1, ..., 9}. The only digit that can be excluded is 0 or 1 or ... or 9. Since we need 9 digits, exactly one digit must be excluded.

Case 1: The digit 0 is excluded. The set of 9 digits used is {1, 2, ..., 9}. The number of 9-digit numbers formed by arranging these 9 distinct non-zero digits is $9!$.

Case 2: A non-zero digit (say 1) is excluded. The set of 9 digits used is {0, 2, 3, ..., 9}. We need to form a 9-digit number using these digits. The total number of permutations of these 9 digits is $9!$. However, permutations starting with 0 do not form a 9-digit number. The number of permutations starting with 0 is the number of ways to arrange the remaining 8 digits (excluding 0 and 1), which is $8!$. So, the number of 9-digit numbers in this case is $9! - 8!$.

There are 9 such cases where a non-zero digit is excluded (excluding 1, excluding 2, ..., excluding 9).

This method is more complicated due to the restriction on the first digit being non-zero.

Let's stick with Method 1 as it directly accounts for the first digit constraint.

Total number of 9-digit numbers with all different digits $= 9 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 = 9 \times 9!$.

Comparing this with the given options:

(A) 10 !

(B) 9 !

(C) 9 $\times$ 9 !

(D) 10 $\times$10 !

The correct option is (C).

The final answer is $\boxed{9 \times 9!}$.

Given:

The word is ‘ARTICLE’.

The letters are A, R, T, I, C, L, E.

Total number of letters = 7.

We need to form words using all these 7 letters.

Constraint: Vowels must occupy the even places.

To Find:

The number of words that can be formed such that vowels occupy the even places.

Solution:

First, identify the vowels and consonants in the word ‘ARTICLE’.

Vowels: A, I, E (There are 3 vowels).

Consonants: R, T, C, L (There are 4 consonants).

Total letters = 3 vowels + 4 consonants = 7.

The word has 7 positions for the letters. Let's number the positions from 1 to 7.

Positions: 1st, 2nd, 3rd, 4th, 5th, 6th, 7th.

The even places are the 2nd, 4th, and 6th positions.

There are 3 even places and 3 vowels.

The odd places are the 1st, 3rd, 5th, and 7th positions.

There are 4 odd places and 4 consonants.

Step 1: Arrange the vowels in the even places.

We have 3 distinct vowels (A, I, E) to be arranged in the 3 distinct even places (2nd, 4th, 6th).

The number of ways to arrange 3 distinct items in 3 distinct positions is $3!$.

Number of ways to arrange vowels $= 3! = 3 \times 2 \times 1 = 6$.

Step 2: Arrange the consonants in the odd places.

We have 4 distinct consonants (R, T, C, L) to be arranged in the 4 distinct odd places (1st, 3rd, 5th, 7th).

The number of ways to arrange 4 distinct items in 4 distinct positions is $4!$.

Number of ways to arrange consonants $= 4! = 4 \times 3 \times 2 \times 1 = 24$.

Since the arrangement of vowels and the arrangement of consonants are independent, the total number of words formed is the product of the number of ways for each step.

Total number of words = (Number of ways to arrange vowels) $\times$ (Number of ways to arrange consonants)

Total number of words $= 3! \times 4! = 6 \times 24 = 144$.

The number of words formed is 144.

Comparing this with the given options:

(A) 1440

(B) 144

(C) 7!

(D) $\binom{4}{4} \times \binom{3}{3}$

Option (D) is $\binom{4}{4} \times \binom{3}{3} = 1 \times 1 = 1$, which is incorrect.

7! = 5040.

The correct option is (B).

The final answer is $\boxed{144}$.

Given:

Number of different green dyes = 5.

Number of different blue dyes = 4.

Number of different red dyes = 3.

We need to choose a combination of dyes.

Constraint: At least one green dye and at least one blue dye must be chosen.

To Find:

The number of combinations of dyes that can be chosen under the given constraints.

Solution:

The selection of dyes from each color group is independent.

For the green dyes, we can choose any number of green dyes from 1 up to 5 (since we need at least one green dye). The number of ways to choose at least one green dye from 5 different green dyes is the total number of ways to choose green dyes (including choosing 0) minus the number of ways to choose no green dyes.

Total ways to choose green dyes from 5 = $\binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 2^5$.

Number of ways to choose no green dyes = $\binom{5}{0} = 1$.

Number of ways to choose at least one green dye $= 2^5 - \binom{5}{0} = 32 - 1 = 31$.

For the blue dyes, we can choose any number of blue dyes from 1 up to 4 (since we need at least one blue dye).

Total ways to choose blue dyes from 4 = $\binom{4}{0} + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 2^4$.

Number of ways to choose no blue dyes = $\binom{4}{0} = 1$.

Number of ways to choose at least one blue dye $= 2^4 - \binom{4}{0} = 16 - 1 = 15$.

For the red dyes, there is no restriction mentioned in the constraints (other than that we are making a choice of dyes). We can choose any number of red dyes from 0 up to 3.

Total ways to choose red dyes from 3 = $\binom{3}{0} + \binom{3}{1} + \binom{3}{2} + \binom{3}{3} = 2^3$.

Number of ways to choose red dyes $= 2^3 = 8$.

Since the selection from each color group is independent, the total number of combinations of dyes satisfying the constraints is the product of the number of ways for each color.

Total number of combinations = (Ways to choose at least one green) $\times$ (Ways to choose at least one blue) $\times$ (Ways to choose red)

Total number of combinations $= (2^5 - 1) \times (2^4 - 1) \times 2^3$

Total number of combinations $= 31 \times 15 \times 8$.

$31 \times 15 = 31 \times (10 + 5) = 310 + 155 = 465$.

$465 \times 8 = (400 + 60 + 5) \times 8 = 3200 + 480 + 40 = 3720$.

The number of combinations of dyes which can be chosen taking at least one green and one blue dye is 3720.

Comparing this with the given options:

(A) 3600

(B) 3720

(C) 3800

(D) 3600

The correct option is (B).

The final answer is $\boxed{3720}$.

Given:

To Find:

The value of $r$.

Solution:

We know the relationship between permutations and combinations:

Substitute the given values from (i) and (ii) into equation (iii):

To find $r!$, divide 840 by 35:

Let's calculate the value:

$\frac{840}{35} = \frac{840 \div 5}{35 \div 5} = \frac{168}{7} = 24$

Now we need to find the integer $r$ whose factorial is 24.

$1! = 1$

$2! = 2 \times 1 = 2$

$3! = 3 \times 2 \times 1 = 6$

$4! = 4 \times 3 \times 2 \times 1 = 24$

Since $4! = 24$, the value of $r$ is 4.

The final answer is $\boxed{4}$.

Given Expression:

To Simplify:

The given expression.

Solution:

We use the identity $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$.

Consider the first two terms: $\binom{15}{8} + \binom{15}{9}$. Here $n=15$, $k=8$, so $k+1=9$.

Now consider the last two terms: $-\binom{15}{6} - \binom{15}{7} = -(\binom{15}{6} + \binom{15}{7})$.

Using the identity $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$ with $n=15$, $k=6$, so $k+1=7$.

So, the expression becomes: $\binom{16}{9} - \binom{16}{7}$.

We also use the property $\binom{n}{k} = \binom{n}{n-k}$.

$\binom{16}{9} = \binom{16}{16-9} = \binom{16}{7}$.

Substitute this into the expression:

$\binom{16}{9} - \binom{16}{7} = \binom{16}{7} - \binom{16}{7} = 0$.

Alternatively, consider the property $\binom{n}{k} = \binom{n}{n-k}$ for the original expression.

$\binom{15}{8} = \binom{15}{15-8} = \binom{15}{7}$.

$\binom{15}{9} = \binom{15}{15-9} = \binom{15}{6}$.

Substitute these back into the given expression:

$\binom{15}{7} + \binom{15}{6} - \binom{15}{6} - \binom{15}{7}$

Rearranging the terms:

$(\binom{15}{7} - \binom{15}{7}) + (\binom{15}{6} - \binom{15}{6}) = 0 + 0 = 0$.

The value of the expression is 0.

The final answer is $\boxed{0}$.

Given:

We are considering permutations of $n$ different objects, taken $r$ at a time.

Repetitions are allowed.

To Find:

The number of such permutations.

Solution:

We need to fill $r$ positions using $n$ different objects, and each object can be used multiple times (repetitions are allowed).

Consider the $r$ positions:

For the first position, there are $n$ choices (any of the $n$ objects).

For the second position, there are still $n$ choices (since repetition is allowed).

For the third position, there are still $n$ choices.

... and so on, up to the $r$-th position.

For the $r$-th position, there are $n$ choices.

By the multiplication principle, the total number of permutations is the product of the number of choices for each position.

Total number of permutations $= n \times n \times n \times ... \times n$ ($r$ times)

Total number of permutations $= n^r$.

The number of permutations of $n$ different objects, taken $r$ at a line, when repetitions are allowed, is $n^r$.

The final answer is $\boxed{n^r}$.

The word is ARTICLE, which has 7 distinct letters: 3 vowels (A, I, E) and 4 consonants (R, T, C, L).

The 7 positions in the word are 1, 2, 3, 4, 5, 6, 7.

The even places are positions 2, 4, and 6. There are 3 even places.

The odd places are positions 1, 3, 5, and 7. There are 4 odd places.

Vowels must occupy the 3 even places. The 3 distinct vowels can be arranged in the 3 distinct even places in $3!$ ways.

Number of ways to arrange vowels $= 3! = 6$.

The 4 consonants must occupy the remaining 4 odd places. The 4 distinct consonants can be arranged in the 4 distinct odd places in $4!$ ways.

Number of ways to arrange consonants $= 4! = 24$.

The total number of words formed is the product of the number of ways to arrange the vowels and the number of ways to arrange the consonants.

Total number of words $= 3! \times 4! = 6 \times 24 = 144$.

The number of words is 144.

Comparing with the options, the correct answer is (B).

The final answer is $\boxed{144}$.

Given:

A bag contains: 5 red, 4 white, and 3 black balls. Total = 12 balls.

We draw 3 balls.

Constraint: At least 2 balls are red.

To Find:

The number of ways to draw 3 balls with at least 2 red balls.

Solution:

The condition "at least 2 are red" means we can have exactly 2 red balls or exactly 3 red balls in the draw of 3 balls.

Case 1: Exactly 2 red balls are drawn.

If 2 balls are red, the remaining $3 - 2 = 1$ ball must be from the non-red balls. The non-red balls are white and black, totaling $4 + 3 = 7$ balls.

Number of ways to choose 2 red balls from 5 $= \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10$.

Number of ways to choose 1 non-red ball from 7 $= \binom{7}{1} = 7$.

Number of ways for Case 1 $= \binom{5}{2} \times \binom{7}{1} = 10 \times 7 = 70$.

Case 2: Exactly 3 red balls are drawn.

If 3 balls are red, the remaining $3 - 3 = 0$ balls must be from the non-red balls (7 non-red balls).

Number of ways to choose 3 red balls from 5 $= \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.

Number of ways to choose 0 non-red balls from 7 $= \binom{7}{0} = 1$.

Number of ways for Case 2 $= \binom{5}{3} \times \binom{7}{0} = 10 \times 1 = 10$.

The total number of ways to draw 3 balls with at least 2 red balls is the sum of the ways in Case 1 and Case 2, as these cases are mutually exclusive.

Total ways = Ways (Case 1) + Ways (Case 2)

Total ways $= 70 + 10 = 80$.

The number of ways in which this can be done if at least 2 are red is 80.

The final answer is $\boxed{80}$.

Given:

We need to form a six-digit number.

All digits of the number must be odd.

The odd digits are {1, 3, 5, 7, 9}. There are 5 distinct odd digits.

To Find:

The number of such six-digit numbers.

Solution:

A six-digit number has 6 positions ($d_1 d_2 d_3 d_4 d_5 d_6$).

For a number to be a six-digit number, the first digit ($d_1$) cannot be 0. However, since we are only using odd digits, and 0 is not an odd digit, the first digit will automatically be non-zero.

For each of the six positions, the digit must be chosen from the set of odd digits {1, 3, 5, 7, 9}.

Unless specified otherwise, repetitions of digits are allowed in number formation problems.

Number of choices for the first digit ($d_1$) = 5 (any of the odd digits).

Number of choices for the second digit ($d_2$) = 5 (any of the odd digits).

Number of choices for the third digit ($d_3$) = 5 (any of the odd digits).

Number of choices for the fourth digit ($d_4$) = 5 (any of the odd digits).

Number of choices for the fifth digit ($d_5$) = 5 (any of the odd digits).

Number of choices for the sixth digit ($d_6$) = 5 (any of the odd digits).

By the multiplication principle, the total number of such six-digit numbers is the product of the number of choices for each position.

Total number of six-digit numbers = $5 \times 5 \times 5 \times 5 \times 5 \times 5 = 5^6$.

Calculate $5^6$:

$5^6 = 5 \times 5 \times 5 \times 5 \times 5 \times 5 = (5^3) \times (5^3) = 125 \times 125$.

$125 \times 125 = (100 + 25) \times (100 + 25) = 10000 + 2 \times 100 \times 25 + 25^2 = 10000 + 5000 + 625 = 15625$.

The number of six-digit numbers, all digits of which are odd, is 15625.

The final answer is $\boxed{15625}$.

Given:

Total number of matches played in a football championship = 153.

Every two teams played one match with each other.

To Find:

The number of teams participating in the championship.

Solution:

Let the number of teams participating in the championship be $n$.

Since every two teams played one match with each other, the total number of matches played is equal to the number of ways to choose 2 teams from the $n$ teams, without regard to the order (since the match between Team A and Team B is the same match as between Team B and Team A).

This is a combination problem. The number of ways to choose 2 teams from $n$ is given by $\binom{n}{2}$.

$\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)(n-2)!}{2 \times 1 \times (n-2)!} = \frac{n(n-1)}{2}$.

We are given that the total number of matches is 153.

So, we have the equation:

Multiply both sides by 2:

$n(n-1) = 153 \times 2$

$n(n-1) = 306$

This is a quadratic equation in $n$. Expand and rearrange:

$n^2 - n = 306$

We need to solve this quadratic equation for $n$. We are looking for two consecutive integers whose product is 306. We can factor 306.

$306 = 1 \times 306$

$306 = 2 \times 153$

$306 = 3 \times 102$

$306 = 6 \times 51$

$306 = 9 \times 34$

$306 = 17 \times 18$

The consecutive integers are 17 and 18. Since $n(n-1) = 306$, and we assume $n$ is positive, $n$ must be the larger of the two consecutive integers. So, $n=18$ and $n-1=17$.

Let's solve the quadratic equation by factoring:

We need two numbers that multiply to -306 and add to -1. The numbers are 17 and -18.

$(n + 17)(n - 18) = 0$

The possible values for $n$ are:

$n + 17 = 0 \implies n = -17$

$n - 18 = 0 \implies n = 18$

Since the number of teams must be a positive integer, $n = 18$ is the valid solution.

The number of teams participating in the championship is 18.

The final answer is $\boxed{18}$.

Given:

We have six ‘+’ signs and four ‘–’ signs.

We need to arrange these 10 signs in a line.

Constraint: No two ‘–’ signs occur together.

To Find:

The total number of ways to arrange the signs such that no two ‘–’ signs are together.

Solution:

To ensure that no two ‘–’ signs occur together, we can first arrange the ‘+’ signs and then place the ‘–’ signs in the spaces created between the ‘+’ signs and at the ends.

Step 1: Arrange the ‘+’ signs.

We have six identical ‘+’ signs. The number of ways to arrange six identical items in a line is 1.

Let's visualize the arrangement of the six ‘+’ signs:

$+ + + + + +$

Step 2: Create spaces for the ‘–’ signs.

When we arrange the 6 ‘+’ signs, there are $6 + 1 = 7$ possible positions where the ‘–’ signs can be placed so that no two ‘–’ signs are adjacent. Let '+' represent a plus sign and '\_' represent a space.

The arrangement of plus signs creates the following spaces:

$\_ + \_ + \_ + \_ + \_ + \_ + \_$

There are 7 available spaces indicated by '\_'.

Step 3: Place the ‘–’ signs in the spaces.

We have four identical ‘–’ signs to place in these 7 available spaces. To ensure no two ‘–’ signs are together, we must place at most one ‘–’ sign in each space. Since we have 4 ‘–’ signs and 7 spaces, we need to choose 4 of the 7 spaces where the ‘–’ signs will be placed.

The number of ways to choose 4 spaces from 7 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Number of ways to choose 4 spaces from 7 $= \binom{7}{4}$.

$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5 \times 4!}{4! \times 3 \times 2 \times 1} = \frac{7 \times 6 \times 5}{6} = 7 \times 5 = 35$.

Once the 4 spaces are chosen, the four identical ‘–’ signs can be placed in these 4 chosen spaces in only 1 way (since they are identical).

The total number of ways to place the ‘–’ signs is the number of ways to choose the spaces multiplied by the number of ways to arrange the identical signs in those spaces (which is 1).

Number of ways to place ‘–’ signs $= \binom{7}{4} \times 1 = 35$.

The total number of ways to arrange the signs such that no two ‘–’ signs are together is the product of the number of ways to arrange the ‘+’ signs and the number of ways to place the ‘–’ signs in the available spaces. Since the arrangement of ‘+’ signs has only 1 way, the total number of ways is simply the number of ways to place the ‘–’ signs.

Total number of arrangements = (Ways to arrange ‘+’) $\times$ (Ways to place ‘–’ in spaces)

Total number of arrangements $= 1 \times 35 = 35$.

The total number of ways is 35.

The final answer is $\boxed{35}$.

Given:

Number of men = 10.

Number of women = 7.

We need to choose a committee of 6 members.

Constraint 1: The committee must include at least 3 men.

Constraint 2: The committee must include at least 2 women.

Constraint 3: Two particular women refuse to serve on the same committee.

To Find:

The number of ways to form the committee under all the given constraints.

Solution:

First, let's find the total number of ways to form the committee of 6 members with at least 3 men and at least 2 women, without considering the constraint about the two particular women.

Let the number of men selected be $m$ and the number of women selected be $w$. We must have $m + w = 6$, with $m \ge 3$ and $w \ge 2$. Also, $m \le 10$ and $w \le 7$.

The possible combinations of $(m, w)$ satisfying $m+w=6$, $m \ge 3$, $w \ge 2$ are:

• $m=3 \implies w = 6-3 = 3$. This satisfies $m \ge 3$ and $w \ge 2$. (3 men, 3 women)
• $m=4 \implies w = 6-4 = 2$. This satisfies $m \ge 3$ and $w \ge 2$. (4 men, 2 women)
• If $m=5$, $w=1$, not $\ge 2$.
• If $m=6$, $w=0$, not $\ge 2$.

So, the valid compositions for the committee are (3 men, 3 women) and (4 men, 2 women).

Number of ways to choose 3 men from 10 and 3 women from 7: $\binom{10}{3} \times \binom{7}{3}$

$\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.

$\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

Ways for (3 men, 3 women) $= 120 \times 35 = 4200$.

Number of ways to choose 4 men from 10 and 2 women from 7: $\binom{10}{4} \times \binom{7}{2}$

$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.

$\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$.

Ways for (4 men, 2 women) $= 210 \times 21 = 4410$.

Total number of committees with at least 3 men and 2 women = $4200 + 4410 = 8610$.

Now, let's consider the constraint that two particular women (let's call them W1 and W2) refuse to serve on the same committee. This means we need to subtract the number of valid committees (meeting the minimum men/women criteria) where both W1 and W2 are members.

If W1 and W2 are on the committee, we have already selected 2 women. The remaining $6 - 2 = 4$ members must be chosen from the remaining $10$ men and $7 - 2 = 5$ women (excluding W1 and W2).

Let $m'$ be the number of men chosen and $w'$ be the number of women chosen from the remaining pools. We need $m' + w' = 4$.

The original constraints were $m \ge 3$ and $w \ge 2$. In this case, $m = m'$ and $w = w' + 2$. So we need $m' \ge 3$ and $w' + 2 \ge 2 \implies w' \ge 0$. Also $m' \le 10$ and $w' \le 5$.

The possible combinations of $(m', w')$ that sum to 4 and satisfy $m' \ge 3, w' \ge 0$ are:

• $m'=3 \implies w' = 4-3 = 1$. This satisfies $m' \ge 3$ and $w' \ge 0$. (3 men, 1 woman from remaining)
• $m'=4 \implies w' = 4-4 = 0$. This satisfies $m' \ge 3$ and $w' \ge 0$. (4 men, 0 women from remaining)

Number of ways to choose 3 men from 10 and 1 woman from the remaining 5: $\binom{10}{3} \times \binom{5}{1}$.

$\binom{10}{3} = 120$.

$\binom{5}{1} = 5$.

Ways for (3 men, 1 woman from remaining) $= 120 \times 5 = 600$. (This committee has 3 men and $1+2=3$ women)

Number of ways to choose 4 men from 10 and 0 women from the remaining 5: $\binom{10}{4} \times \binom{5}{0}$.

$\binom{10}{4} = 210$.

$\binom{5}{0} = 1$.

Ways for (4 men, 0 women from remaining) $= 210 \times 1 = 210$. (This committee has 4 men and $0+2=2$ women)

Total number of committees where W1 and W2 are together (and satisfy the minimum requirements) $= 600 + 210 = 810$.

The number of committees where the two particular women refuse to serve together is the total number of valid committees minus the number of valid committees where they are together.

Number of ways = (Total valid committees) - (Valid committees with W1 and W2 together)

Number of ways $= 8610 - 810 = 7800$.

The final answer is $\boxed{7800}$.

Given:

A box contains:

• White balls: 2
• Black balls: 3
• Red balls: 4

Total number of balls $= 2 + 3 + 4 = 9$.

We need to draw 3 balls from the box.

Constraint: At least one black ball must be included in the draw.

To Find:

The number of ways to draw 3 balls such that at least one black ball is included.

Solution:

The condition "at least one black ball" means that in the group of 3 balls drawn, there can be exactly 1 black ball, exactly 2 black balls, or exactly 3 black balls.

The non-black balls available for selection are the white and red balls, totaling $2 + 4 = 6$ balls.

Let's consider each possible case for the number of black balls drawn:

Case 1: Exactly 1 black ball is drawn.

We choose 1 black ball from the 3 available black balls ($\binom{3}{1}$) and $3 - 1 = 2$ non-black balls from the 6 available non-black balls ($\binom{6}{2}$).

Number of ways for Case 1 $= \binom{3}{1} \times \binom{6}{2} = 3 \times \frac{6 \times 5}{2 \times 1} = 3 \times 15 = 45$.

Case 2: Exactly 2 black balls are drawn.

We choose 2 black balls from the 3 available black balls ($\binom{3}{2}$) and $3 - 2 = 1$ non-black ball from the 6 available non-black balls ($\binom{6}{1}$).

Number of ways for Case 2 $= \binom{3}{2} \times \binom{6}{1} = 3 \times 6 = 18$.

Case 3: Exactly 3 black balls are drawn.

We choose 3 black balls from the 3 available black balls ($\binom{3}{3}$) and $3 - 3 = 0$ non-black balls from the 6 available non-black balls ($\binom{6}{0}$).

Number of ways for Case 3 $= \binom{3}{3} \times \binom{6}{0} = 1 \times 1 = 1$.

The total number of ways to draw 3 balls with at least one black ball is the sum of the ways in the three mutually exclusive cases.

Total ways = Ways (Case 1) + Ways (Case 2) + Ways (Case 3)

Total ways $= 45 + 18 + 1 = 64$.

The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is 64.

The final answer is $\boxed{64}$.

The statement is False.

Justification:

Let the total number of points be $n = 12$.

Let the number of collinear points be $m = 5$.

The total number of lines that can be formed by joining any two points out of $n$ points, assuming no three are collinear, is given by ${}^{n}\text{C}_2$. In this case, it would be ${}^{12}\text{C}_2$.

The number of lines that can be formed by joining any two points out of the $m$ collinear points is given by ${}^{m}\text{C}_2$. In this case, it would be ${}^{5}\text{C}_2$.

However, these $m$ collinear points form only one single line.

To find the actual number of lines formed by $n$ points where $m$ of them are collinear, we first calculate the total number of lines as if no three points were collinear (${}^{n}\text{C}_2$). Then, we subtract the lines formed by the $m$ collinear points (${}^{m}\text{C}_2$) because they do not form distinct lines. Finally, we add back 1 for the single line formed by all the $m$ collinear points.

The correct formula for the number of distinct lines is:

${}^{n}\text{C}_2 - {}^{m}\text{C}_2 + 1$

Substituting the given values $n=12$ and $m=5$:

Number of lines $= {}^{12}\text{C}_2 - {}^{5}\text{C}_2 + 1$

Let's calculate the combination values:

${}^{12}\text{C}_2 = \frac{12 \times 11}{2 \times 1} = \frac{132}{2} = 66$

${}^{5}\text{C}_2 = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10$

So, the actual number of lines is:

$66 - 10 + 1 = 56 + 1 = 57$

The given statement claims the number of lines is ${}^{12}\text{C}_2 - {}^{5}\text{C}_2$, which calculates to $66 - 10 = 56$.

Since $57 \neq 56$, the statement is incorrect.

Therefore, the statement "There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ${}^{12}\text{C}_2 – {}^{5}\text{C}_2$" is False.

The statement is False.

Justification:

Let the three letters be $L_1, L_2, L_3$. Let the five letterboxes be $B_1, B_2, B_3, B_4, B_5$.

For the first letter ($L_1$), there are 5 choices of letterboxes it can be placed in ($B_1$ to $B_5$).

For the second letter ($L_2$), there are also 5 choices of letterboxes it can be placed in, irrespective of where $L_1$ was placed (since a letterbox can hold multiple letters).

For the third letter ($L_3$), there are likewise 5 choices of letterboxes.

Since the choices for each letter are independent, the total number of ways to post the three letters is the product of the number of choices for each letter.

Total number of ways $= \text{(Choices for } L_1\text{)} \times \text{(Choices for } L_2\text{)} \times \text{(Choices for } L_3\text{)}$

Total number of ways $= 5 \times 5 \times 5 = 5^3$

Calculating the value:

$5^3 = 5 \times 5 \times 5 = 125$

The statement claims the number of ways is $3^5$.

Calculating the value given in the statement:

$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243$

Since $125 \neq 243$, the statement is incorrect.

The number of ways to post 3 letters in 5 letterboxes is $5^3$, not $3^5$.

Therefore, the statement "Three letters can be posted in five letterboxes in $3^5$ ways" is False.

The statement is False.

Justification:

Let $n$ be the total number of distinct things and we are taking $r$ things at a time to form permutations, where $m$ particular things must always occur together ($m \leq r \leq n$).

To find the number of such permutations, we can follow these steps:

1. Treat the $m$ particular things as a single block. Now we have $n-m$ individual items plus this block, giving a total of $(n-m) + 1$ entities.

2. We need to select $r$ items for the permutation, and these must include the $m$ particular things (as a block). So, we need to select the remaining $r-m$ items from the $n-m$ individual items. The number of ways to choose these $r-m$ items is ${}^{n-m}\text{C}_{r-m}$.

3. Now we have $r-m$ selected individual items and the block of $m$ items. These form a group of $(r-m)+1$ entities that need to be arranged in the $r$ positions of the permutation.

4. Consider the arrangement of these $(r-m+1)$ entities. The block of $m$ items must stay together. There are $(r-m+1)$ possible positions for the block within the $r$ spots (starting at position 1, 2, ..., up to $r-m+1$).

5. The remaining $r-m$ individual items must be arranged in the $r-m$ positions outside the block. There are $(r-m)!$ ways to arrange these $r-m$ items.

6. The $m$ items within the block can be arranged in $m!$ ways.

Combining these steps, the correct number of permutations is:

Number of permutations = (Ways to choose $r-m$ items) $\times$ (Ways to place the block) $\times$ (Ways to arrange items outside block) $\times$ (Ways to arrange items within block)

Number of permutations $= {}^{n-m}\text{C}_{r-m} \times (r-m+1) \times (r-m)! \times m!$

Using the formula for combination ${}^{k}\text{C}_p = \frac{k!}{p!(k-p)!}$ and permutation ${}^{k}\text{P}_p = \frac{k!}{(k-p)!}$:

${}^{n-m}\text{C}_{r-m} = \frac{(n-m)!}{(r-m)!(n-m - (r-m))!} = \frac{(n-m)!}{(r-m)!(n-r)!}$

So, the number of permutations $= \frac{(n-m)!}{(r-m)!(n-r)!} \times (r-m+1) \times (r-m)! \times m!$

We can cancel out $(r-m)!$ from the numerator and denominator:

Number of permutations $= \frac{(n-m)!}{(n-r)!} \times (r-m+1) \times m!$

Recognizing that $\frac{(n-m)!}{(n-r)!} = {}^{n-m}\text{P}_{r-m}$, the correct formula is:

${}^{n-m}\text{P}_{r-m} \times (r-m+1) \times m!$

The statement claims the number of permutations is ${}^{n – m}\text{P}_{r – m} \times {}^{r}\text{P}_{m}$.

Since $(r-m+1) \times m! \neq {}^{r}\text{P}_{m} = \frac{r!}{(r-m)!}$ in general, the formula provided in the statement is incorrect.

Therefore, the statement "In the permutations of n things, r taken together, the number of permutations in which m particular things occur together is ${}^{n – m}\text{P}_{r – m} \times {}^{r}\text{P}_{m}$" is False.

The statement is True.

Justification:

We have 12 stalls to fill, and for each stall, we can choose one animal from three types: horses, cows, or calves.

There are plenty of each type of animal (not less than 12 each), so the supply of animals of any type does not run out as we fill the stalls.

Consider the first stall. There are 3 choices for the animal to be placed in this stall (Horse, Cow, or Calf).

Consider the second stall. There are also 3 choices for the animal, independent of the choice made for the first stall.

This applies to each of the 12 stalls.

For the 1st stall, there are 3 choices.

For the 2nd stall, there are 3 choices.

...

For the 12th stall, there are 3 choices.

Since the choice for each stall is independent, the total number of ways to load the animals into the 12 stalls is the product of the number of choices for each stall.

Total number of ways $= 3 \times 3 \times 3 \times \dots \times 3$ (12 times)

Total number of ways $= 3^{12}$

This matches the number of ways given in the statement.

Therefore, the statement "In a steamer there are stalls for 12 animals, and there are horses, cows and calves (not less than 12 each) ready to be shipped. They can be loaded in $3^{12}$ ways" is True.

The statement is True.

Justification:

We are asked to find the number of combinations when taking some or all of $n$ distinct objects at a time.

This means we can take 1 object, or 2 objects, or 3 objects, ..., or $n$ objects.

The number of ways to choose $k$ objects from $n$ distinct objects is given by the combination formula ${}^{n}\text{C}_k = \frac{n!}{k!(n-k)!}$.

The number of combinations when taking some or all objects is the sum of the number of ways to choose 1, 2, ..., up to $n$ objects:

${}^{n}\text{C}_1 + {}^{n}\text{C}_2 + {}^{n}\text{C}_3 + \dots + {}^{n}\text{C}_n$

We know the binomial expansion:

$(1+x)^n = {}^{n}\text{C}_0 + {}^{n}\text{C}_1 x + {}^{n}\text{C}_2 x^2 + \dots + {}^{n}\text{C}_n x^n$

Setting $x = 1$, we get:

$(1+1)^n = {}^{n}\text{C}_0 + {}^{n}\text{C}_1 (1) + {}^{n}\text{C}_2 (1)^2 + \dots + {}^{n}\text{C}_n (1)^n$

$2^n = {}^{n}\text{C}_0 + {}^{n}\text{C}_1 + {}^{n}\text{C}_2 + \dots + {}^{n}\text{C}_n$

The term ${}^{n}\text{C}_0$ represents the number of ways to choose 0 objects from $n$, which corresponds to taking none of the objects (the empty set). ${}^{n}\text{C}_0 = 1$.

The sum ${}^{n}\text{C}_0 + {}^{n}\text{C}_1 + {}^{n}\text{C}_2 + \dots + {}^{n}\text{C}_n$ represents the total number of subsets of a set with $n$ elements, which is $2^n$.

We are interested in the number of combinations when taking "some or all", which means we exclude the case of taking none (${}^{n}\text{C}_0$).

So, the number of combinations when taking some or all is:

$({}^{n}\text{C}_0 + {}^{n}\text{C}_1 + {}^{n}\text{C}_2 + \dots + {}^{n}\text{C}_n) - {}^{n}\text{C}_0$

Which is $2^n - 1$.

This matches the formula given in the statement.

Therefore, the statement "If some or all of n objects are taken at a time, the number of combinations is $2^n – 1$" is True.

The statement is True.

Justification:

We have 4 identical red balls and 5 identical black balls.

When dealing with identical items, the number of ways to select items of a particular type depends only on the quantity selected, not the specific items themselves.

The number of ways to select red balls from 4 identical red balls is the number of possible counts we can select:

We can select 0, 1, 2, 3, or 4 red balls. This gives $4+1=5$ ways.

The number of ways to select black balls from 5 identical black balls is the number of possible counts we can select:

We can select 0, 1, 2, 3, 4, or 5 black balls. This gives $5+1=6$ ways.

The total number of distinct selections possible from the bag (including the selection of no balls) is the product of the number of ways to select red balls and the number of ways to select black balls.

Total number of selections $= (\text{Ways to select red balls}) \times (\text{Ways to select black balls})$

Total number of selections $= 5 \times 6 = 30$

We are asked for the number of selections containing at least one red ball.

This is equivalent to finding the total number of selections and subtracting the number of selections that contain no red balls.

The number of selections with no red balls means we select 0 red balls (which can be done in 1 way, i.e., choosing a count of 0) and any number of black balls (0 to 5, which is 6 ways).

Number of selections with no red balls $= (\text{Ways to select 0 red balls}) \times (\text{Ways to select black balls})$

Number of selections with no red balls $= 1 \times 6 = 6$

The number of selections with at least one red ball is:

Number of selections (at least one red) = Total number of selections - Number of selections (no red)

Number of selections (at least one red) $= 30 - 6 = 24$

The statement claims that there will be 24 selections containing at least one red ball.

Our calculation yields 24.

Therefore, the statement "There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical" is True.

The statement is True.

Justification:

Given:

Total number of guests = 18.

The table is long, with 9 seats on each of the two sides.

4 particular guests must sit on one specific side (let's call it Side A).

3 other particular guests must sit on the other side (let's call it Side B).

Solution:

Let Side A be the side where the 4 particular guests must sit, and Side B be the side where the 3 particular guests must sit.

Number of seats on Side A = 9.

Number of seats on Side B = 9.

The 4 particular guests occupy 4 seats on Side A. Remaining seats on Side A = $9 - 4 = 5$.

The 3 particular guests occupy 3 seats on Side B. Remaining seats on Side B = $9 - 3 = 6$.

Total number of particular guests fixed = $4 + 3 = 7$.

Number of remaining guests = Total guests - Fixed guests = $18 - 7 = 11$.

These 11 remaining guests must be distributed into the remaining $5 + 6 = 11$ seats on the two sides.

We need to choose 5 guests for the remaining 5 seats on Side A from the 11 remaining guests. The number of ways to do this is ${}^{11}\text{C}_5$.

${}^{11}\text{C}_5 = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!}$.

Once 5 guests are chosen for Side A, the remaining $11 - 5 = 6$ guests must sit on Side B. The number of ways to choose these 6 guests from the remaining 6 is ${}^{6}\text{C}_6 = 1$.

So, the number of ways to distribute the 11 remaining guests between the two sides is ${}^{11}\text{C}_5 = \frac{11!}{5!6!}$.

Now consider the arrangement of guests on each side:

On Side A, there are 4 fixed guests plus the 5 chosen guests, totaling $4 + 5 = 9$ guests. These 9 guests can be arranged in the 9 seats on Side A in $9!$ ways.

On Side B, there are 3 fixed guests plus the 6 guests who were not chosen for Side A, totaling $3 + 6 = 9$ guests. These 9 guests can be arranged in the 9 seats on Side B in $9!$ ways.

The total number of ways to make the seating arrangements is the product of the number of ways to distribute the guests and the number of ways to arrange them on each side.

Total number of arrangements = (Ways to distribute remaining guests) $\times$ (Ways to arrange guests on Side A) $\times$ (Ways to arrange guests on Side B)

Total number of arrangements $= {}^{11}\text{C}_5 \times 9! \times 9!$

Total number of arrangements $= \frac{11!}{5!6!} \times 9! \times 9!$

This matches the formula given in the statement.

Therefore, the statement "The number of ways in which the seating arrangements can be made is $\frac{11!}{5!\;6!} (9!) (9!)$" is True.

The statement is False.

Justification:

Let the two groups of questions be Group A and Group B.

Number of questions in Group A = 6.

Number of questions in Group B = 6.

Total number of questions = $6 + 6 = 12$.

The candidate must answer a total of 7 questions.

Let $q_A$ be the number of questions chosen from Group A and $q_B$ be the number of questions chosen from Group B.

We must have $q_A + q_B = 7$.

The constraint is that the candidate is not permitted to attempt more than 5 questions from either group.

So, $q_A \leq 5$ and $q_B \leq 5$.

Also, $q_A$ and $q_B$ must be non-negative integers and cannot exceed the total number of questions in their respective groups (i.e., $0 \leq q_A \leq 6$ and $0 \leq q_B \leq 6$).

We need to find pairs $(q_A, q_B)$ such that $q_A + q_B = 7$ and $q_A \leq 5$, $q_B \leq 5$, $0 \leq q_A \leq 6$, $0 \leq q_B \leq 6$.

Let's list the possible pairs $(q_A, q_B)$ that sum to 7 and check the constraints:

- If $q_A = 0$, then $q_B = 7$. This violates $q_B \leq 5$. (Invalid)

- If $q_A = 1$, then $q_B = 6$. This violates $q_B \leq 5$. (Invalid)

- If $q_A = 2$, then $q_B = 5$. This satisfies $q_A \leq 5$ and $q_B \leq 5$. (Valid)

- If $q_A = 3$, then $q_B = 4$. This satisfies $q_A \leq 5$ and $q_B \leq 5$. (Valid)

- If $q_A = 4$, then $q_B = 3$. This satisfies $q_A \leq 5$ and $q_B \leq 5$. (Valid)

- If $q_A = 5$, then $q_B = 2$. This satisfies $q_A \leq 5$ and $q_B \leq 5$. (Valid)

- If $q_A = 6$, then $q_B = 1$. This violates $q_A \leq 5$. (Invalid)

- If $q_A = 7$, then $q_B = 0$. This violates $q_A \leq 5$. (Invalid)

So, the possible ways to choose the 7 questions are:

1. Choose 2 questions from Group A and 5 questions from Group B.

2. Choose 3 questions from Group A and 4 questions from Group B.

3. Choose 4 questions from Group A and 3 questions from Group B.

4. Choose 5 questions from Group A and 2 questions from Group B.

We calculate the number of ways for each case using combinations:

Case 1: Ways to choose 2 from 6 (Group A) and 5 from 6 (Group B) is ${}^{6}\text{C}_2 \times {}^{6}\text{C}_5$.

${}^{6}\text{C}_2 = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$

${}^{6}\text{C}_5 = \frac{6!}{5!(6-5)!} = \frac{6!}{5!1!} = 6$

Number of ways for Case 1 = $15 \times 6 = 90$.

Case 2: Ways to choose 3 from 6 (Group A) and 4 from 6 (Group B) is ${}^{6}\text{C}_3 \times {}^{6}\text{C}_4$.

${}^{6}\text{C}_3 = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

${}^{6}\text{C}_4 = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$

Number of ways for Case 2 = $20 \times 15 = 300$.

Case 3: Ways to choose 4 from 6 (Group A) and 3 from 6 (Group B) is ${}^{6}\text{C}_4 \times {}^{6}\text{C}_3$.

${}^{6}\text{C}_4 = 15$

${}^{6}\text{C}_3 = 20$

Number of ways for Case 3 = $15 \times 20 = 300$.

Case 4: Ways to choose 5 from 6 (Group A) and 2 from 6 (Group B) is ${}^{6}\text{C}_5 \times {}^{6}\text{C}_2$.

${}^{6}\text{C}_5 = 6$

${}^{6}\text{C}_2 = 15$

Number of ways for Case 4 = $6 \times 15 = 90$.

The total number of ways to choose the 7 questions is the sum of the ways for these valid cases.

Total number of ways = (Ways for Case 1) + (Ways for Case 2) + (Ways for Case 3) + (Ways for Case 4)

Total number of ways = $90 + 300 + 300 + 90 = 780$.

The statement claims that the candidate can choose the seven questions in 650 ways.

Our calculation shows the number of ways is 780.

Since $780 \neq 650$, the statement is incorrect.

Therefore, the statement "He can choose the seven questions in 650 ways" is False.

The statement is False.

Justification:

Total number of vacancies = 12.

Total number of candidates = 25.

Number of Scheduled Caste (SC) candidates = 5.

Number of other candidates (Non-SC) = $25 - 5 = 20$.

Number of vacancies reserved for SC candidates = 3.

Number of remaining vacancies (open to all) = $12 - 3 = 9$.

The selection process consists of two parts:

1. Selecting candidates for the reserved vacancies.

2. Selecting candidates for the open vacancies.

Step 1: Select 3 SC candidates for the 3 reserved vacancies.

These 3 candidates must be chosen from the 5 available SC candidates.

Number of ways to select 3 SC candidates = ${}^{5}\text{C}_3$.

${}^{5}\text{C}_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.

Step 2: Select 9 candidates for the 9 open vacancies.

These vacancies are open to all remaining candidates.

After selecting 3 SC candidates, the number of remaining SC candidates is $5 - 3 = 2$.

The number of other candidates (Non-SC) is still 20.

Total number of candidates available for the open vacancies = (Remaining SC candidates) + (Other candidates) = $2 + 20 = 22$.

We need to choose 9 candidates from these 22 available candidates for the 9 open vacancies.

Number of ways to select 9 candidates for open vacancies = ${}^{22}\text{C}_9$.

The total number of ways to make the selection is the product of the ways in Step 1 and Step 2.

Total number of ways = (Ways to select SC for reserved) $\times$ (Ways to select from remaining for open)

Total number of ways $= {}^{5}\text{C}_3 \times {}^{22}\text{C}_9$.

The statement claims the number of ways is ${}^{5}\text{C}_3 \times {}^{20}\text{C}_9$.

The formula in the statement incorrectly assumes that the 9 open vacancies can only be filled by the 20 non-SC candidates. It ignores the 2 SC candidates who were not selected for the reserved vacancies, who are also eligible for the open vacancies.

The correct number of ways is ${}^{5}\text{C}_3 \times {}^{22}\text{C}_9$, not ${}^{5}\text{C}_3 \times {}^{20}\text{C}_9$.

Therefore, the statement "the number of ways in which the selection can be made is ${}^{5}\text{C}_3 \times {}^{20}\text{C}_9$" is False.

Let M be the number of Mathematics books, P be the number of Physics books, and E be the number of English books.

Given:

Number of Mathematics books, $n(M) = 3$

Number of Physics books, $n(P) = 4$

Number of English books, $n(E) = 5$

(a) One book of each subject:

To choose one book of each subject, we need to select 1 book from Mathematics, 1 book from Physics, and 1 book from English.

Number of ways to select 1 Mathematics book from 3 is $\binom{3}{1}$.

Number of ways to select 1 Physics book from 4 is $\binom{4}{1}$.

Number of ways to select 1 English book from 5 is $\binom{5}{1}$.

The total number of different collections is the product of the number of ways to select from each subject.

Total collections $= \binom{3}{1} \times \binom{4}{1} \times \binom{5}{1}$

$= 3 \times 4 \times 5$

$= 60$

This matches option (ii) in column $C_2$.

(b) At least one book of each subject:

The number of ways to select at least one book from a set of $n$ distinct books is $2^n - 1$.

Number of ways to select at least one Mathematics book from 3 is $2^3 - 1 = 8 - 1 = 7$.

Number of ways to select at least one Physics book from 4 is $2^4 - 1 = 16 - 1 = 15$.

Number of ways to select at least one English book from 5 is $2^5 - 1 = 32 - 1 = 31$.

To form a collection with at least one book from each subject, we multiply the number of ways to select at least one from each subject.

Total collections $= (2^3 - 1) \times (2^4 - 1) \times (2^5 - 1)$

$= 7 \times 15 \times 31$

$= 105 \times 31$

$= 3255$

This matches option (iii) in column $C_2$.

(c) At least one book of English:

This means we must select at least one English book, and we can select any number of books (including zero) from Mathematics and Physics.

Number of ways to select at least one English book from 5 is $2^5 - 1 = 32 - 1 = 31$.

Number of ways to select any number of Mathematics books from 3 is $2^3 = 8$.

Number of ways to select any number of Physics books from 4 is $2^4 = 16$.

To form a collection with at least one English book and any number of books from other subjects, we multiply the number of ways to select from each subject category.

Total collections $= (2^5 - 1) \times 2^3 \times 2^4$

$= 31 \times 8 \times 16$

$= 31 \times 128$

$= 3968$

This matches option (i) in column $C_2$.

Matching:

(a) $\to$ (ii)

(b) $\to$ (iii)

(c) $\to$ (i)

Given:

Number of boys = 5

Number of girls = 5

Total number of people = $5 + 5 = 10$

We need to find the number of ways to arrange these 10 people in a line under different conditions.

(a) Boys and girls alternate:

Since there are equal numbers of boys and girls, the arrangement must alternate starting with a boy or a girl.

Case 1: Boy, Girl, Boy, Girl, ... (B G B G B G B G B G)

The 5 boys can be arranged in the 5 odd positions in $5!$ ways.

The 5 girls can be arranged in the 5 even positions in $5!$ ways.

Number of arrangements in Case 1 = $5! \times 5!$.

Case 2: Girl, Boy, Girl, Boy, ... (G B G B G B G B G B)

The 5 girls can be arranged in the 5 odd positions in $5!$ ways.

The 5 boys can be arranged in the 5 even positions in $5!$ ways.

Number of arrangements in Case 2 = $5! \times 5!$.

Total number of alternating arrangements = (Arrangements in Case 1) + (Arrangements in Case 2)

Total arrangements = $5! \times 5! + 5! \times 5! = 2 \times (5! \times 5!) = 2! \times 5! \times 5!$.

This matches option (iv) in column $C_2$. Note that this is also equal to $(5!)^2 + (5!)^2$, which is option (iii).

(b) No two girls sit together:

To ensure no two girls sit together, we must place the boys first and then place the girls in the spaces between the boys or at the ends.

First, arrange the 5 boys. The number of ways to arrange 5 boys is $5!$.

When the boys are arranged, there are 6 possible positions (indicated by underscores) where the girls can be placed:

_ B _ B _ B _ B _ B _

We need to choose 5 of these 6 positions for the 5 girls. The number of ways to choose 5 positions from 6 is $\binom{6}{5}$.

Once the 5 positions are chosen, the 5 girls can be arranged in these positions in $5!$ ways.

Total number of arrangements = (Ways to arrange boys) $\times$ (Ways to choose positions for girls) $\times$ (Ways to arrange girls in chosen positions)

Total arrangements = $5! \times \binom{6}{5} \times 5!$

Since $\binom{6}{5} = 6$,

Total arrangements = $5! \times 6 \times 5! = (6 \times 5!) \times 5! = 6! \times 5!$.

This matches option (i) in column $C_2$.

(c) All the girls sit together:

Treat the group of 5 girls as a single unit or block.

Now, we are arranging the 5 boys and this single block of girls. This gives a total of $5 + 1 = 6$ entities to arrange.

The number of ways to arrange these 6 entities is $6!$.

Within the block of girls, the 5 girls can be arranged among themselves in $5!$ ways.

Total number of arrangements = (Ways to arrange the 6 entities) $\times$ (Ways to arrange girls within the block)

Total arrangements = $6! \times 5!$.

This matches option (i) in column $C_2$.

(d) All the girls are never together:

The total number of ways to arrange 10 people (5 boys and 5 girls) without any restrictions is $10!$.

The number of ways where all the girls *do* sit together was calculated in part (c), which is $6! \times 5!$.

The number of ways where all the girls are never together is the total number of arrangements minus the number of arrangements where all the girls are together.

Total arrangements - Arrangements where all girls sit together $= 10! - (6! \times 5!)$.

This matches option (ii) in column $C_2$.

Matching:

(a) Boys and girls alternate: $\to$ (iv) $2! \ 5! \ 5!$

(b) No two girls sit together : $\to$ (i) $5! \times 6!$

(c) All the girls sit together : $\to$ (i) $5! \times 6!$

(d) All the girls are never together : $\to$ (ii) $10! \ – \ 5! \ 6!$

Note: Conditions (b) and (c) result in the same number of arrangements, which both match option (i) in column $C_2$. Also, options (iii) and (iv) in $C_2$ are numerically equal ($2 \times (5!)^2$). Based on the structure $2! \times 5! \times 5!$ better represents the two cases for alternating arrangements.

Given:

Total number of professors = 10

Total number of lecturers = 20

Committee size required: 2 professors and 3 lecturers.

(a) In how many ways can the committee be formed?

We need to choose 2 professors out of 10 and 3 lecturers out of 20.

Number of ways to choose 2 professors from 10 is $\binom{10}{2}$.

Number of ways to choose 3 lecturers from 20 is $\binom{20}{3}$.

The total number of ways to form the committee is the product of the number of ways to choose professors and lecturers.

Total ways $= \binom{10}{2} \times \binom{20}{3}$

This matches option (iv) in column $C_2$.

(b) In how many ways can the committee be formed if a particular professor is included?

If a particular professor must be included, we have already selected 1 professor.

We need to select $2 - 1 = 1$ more professor from the remaining $10 - 1 = 9$ professors.

Number of ways to choose the remaining professor is $\binom{9}{1}$.

We still need to choose 3 lecturers from the 20 lecturers.

Number of ways to choose 3 lecturers is $\binom{20}{3}$.

The total number of ways is the product of the number of ways to choose the remaining professor and the lecturers.

Total ways $= \binom{9}{1} \times \binom{20}{3}$

This matches option (iii) in column $C_2$.

(c) In how many ways can the committee be formed if a particular lecturer is included?

If a particular lecturer must be included, we have already selected 1 lecturer.

We need to select $3 - 1 = 2$ more lecturers from the remaining $20 - 1 = 19$ lecturers.

Number of ways to choose the remaining lecturers is $\binom{19}{2}$.

We still need to choose 2 professors from the 10 professors.

Number of ways to choose 2 professors is $\binom{10}{2}$.

The total number of ways is the product of the number of ways to choose the professors and the remaining lecturers.

Total ways $= \binom{10}{2} \times \binom{19}{2}$

This matches option (ii) in column $C_2$.

(d) In how many ways can the committee be formed if a particular lecturer is excluded?

If a particular lecturer must be excluded, we need to choose the 3 lecturers from the remaining $20 - 1 = 19$ lecturers.

Number of ways to choose 3 lecturers is $\binom{19}{3}$.

We still need to choose 2 professors from the 10 professors.

Number of ways to choose 2 professors is $\binom{10}{2}$.

The total number of ways is the product of the number of ways to choose the professors and the lecturers from the reduced pool.

Total ways $= \binom{10}{2} \times \binom{19}{3}$

This matches option (i) in column $C_2$.

Matching:

(a) $\to$ (iv)

(b) $\to$ (iii)

(c) $\to$ (ii)

(d) $\to$ (i)

Given digits: $\{1, 2, 3, 4, 5, 6, 7\}$. Total number of distinct digits = 7.

We are forming 4-digit numbers with different digits.

(a) How many numbers are formed?

We need to choose 4 distinct digits from the 7 available digits and arrange them in 4 positions.

This is the number of permutations of 7 distinct objects taken 4 at a time, denoted by $P(7, 4)$ or $^{7}P_4$.

$P(7, 4) = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840$.

Number of numbers formed = 840.

This matches option (i) in column $C_2$.

(b) How many numbers are exactly divisible by 2?

A number is divisible by 2 if its units digit is even.

The even digits in the set $\{1, 2, 3, 4, 5, 6, 7\}$ are $\{2, 4, 6\}$. There are 3 choices for the units digit.

Let the 4-digit number be ABCD.

The units digit (D) can be 2, 4, or 6 (3 choices).

The remaining 3 digits (A, B, C) must be chosen and arranged from the remaining $7 - 1 = 6$ digits.

The number of ways to choose and arrange 3 digits from 6 is $P(6, 3) = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 \times 5 \times 4 = 120$.

Total number of numbers divisible by 2 = (Number of choices for units digit) $\times$ (Number of arrangements of the remaining 3 digits)

Total numbers = $3 \times 120 = 360$.

This matches option (iii) in column $C_2$.

(c) How many numbers are exactly divisible by 25?

A number is divisible by 25 if the number formed by its last two digits is divisible by 25.

The possible two-digit endings divisible by 25 are 00, 25, 50, 75, etc.

Using distinct digits from $\{1, 2, 3, 4, 5, 6, 7\}$, the only possible two-digit endings are 25 and 75.

Case 1: The number ends with 25 (CD = 25).

The digits used are 2 and 5. We need to choose and arrange the first two digits (A and B) from the remaining $7 - 2 = 5$ digits.

Number of ways to arrange the first two digits = $P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$.

Case 2: The number ends with 75 (CD = 75).

The digits used are 7 and 5. We need to choose and arrange the first two digits (A and B) from the remaining $7 - 2 = 5$ digits.

Number of ways to arrange the first two digits = $P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$.

Total number of numbers divisible by 25 = (Numbers ending in 25) + (Numbers ending in 75) = $20 + 20 = 40$.

This matches option (iv) in column $C_2$.

(d) How many of these are exactly divisible by 4?

A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

We need to find two-digit numbers formed using distinct digits from $\{1, 2, 3, 4, 5, 6, 7\}$ that are divisible by 4.

Possible two-digit endings (CD) that are divisible by 4 and use distinct digits from the set:

12, 16, 24, 32, 36, 52, 56, 64, 72, 76.

There are 10 such possible endings.

For each ending (CD), the two digits are used. We need to choose and arrange the remaining 2 digits for the first two positions (A and B) from the remaining $7 - 2 = 5$ digits.

Number of ways to arrange the first two digits = $P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$.

Total number of numbers divisible by 4 = (Number of possible endings CD) $\times$ (Number of arrangements of the first two digits)

Total numbers = $10 \times 20 = 200$.

This matches option (ii) in column $C_2$.

Matching:

(a) How many numbers are formed? $\to$ (i) 840

(b) How many numbers are exactly divisible by 2? $\to$ (iii) 360

(c) How many numbers are exactly divisible by 25? $\to$ (iv) 40

(d) How many of these are exactly divisible by 4? $\to$ (ii) 200

Given word: MONDAY

The word MONDAY has 6 distinct letters: M, O, N, D, A, Y.

Number of letters = 6.

Vowels in MONDAY are O, A. Number of vowels = 2.

Consonants in MONDAY are M, N, D, Y. Number of consonants = 4.

(a) 4 letters are used at a time:

We need to form a 4-letter word using 4 distinct letters from the 6 available letters.

This is a permutation problem, as the order of letters matters.

The number of permutations of $n$ distinct objects taken $r$ at a time is given by $P(n, r) = \frac{n!}{(n-r)!}$.

Here, $n = 6$ (total letters) and $r = 4$ (letters used).

Number of words $= P(6, 4) = \frac{6!}{(6-4)!} = \frac{6!}{2!}$

$= \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 6 \times 5 \times 4 \times 3 = 360$.

Number of words formed = 360.

This matches option (iii) in column $C_2$.

(b) All letters are used at a time:

We need to form a 6-letter word using all 6 distinct letters.

This is the number of permutations of 6 distinct objects, which is $6!$ or $P(6, 6)$.

Number of words $= 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

Number of words formed = 720.

This matches option (i) in column $C_2$.

(c) All letters are used but the first is a vowel:

We are forming a 6-letter word using all 6 distinct letters.

The first position must be occupied by a vowel.

There are 2 vowels (O, A) in the word MONDAY.

Number of choices for the first letter (which must be a vowel) = 2.

After choosing the first letter, there are $6 - 1 = 5$ remaining letters.

These 5 remaining letters can be arranged in the remaining 5 positions in $5!$ ways.

Number of arrangements for the remaining 5 letters = $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Total number of words = (Number of choices for the first letter) $\times$ (Number of arrangements of the remaining letters)

Total words $= 2 \times 120 = 240$.

Number of words formed = 240.

This matches option (ii) in column $C_2$.

Matching:

(a) 4 letters are used at a time $\to$ (iii) 360

(b) All letters are used at a time $\to$ (i) 720

(c) All letters are used but the first is a vowel $\to$ (ii) 240