Class 11- Applied Mathematics-NCERT Handbook Solutions-Chapter-8 (Combinatorics) -

Class 11 Applied Mathematics - Chapter 8 Solutions

Check Your Progress 8.1

4! + 5! = 9!

Solution:

Step 1: Calculate the Left Hand Side (LHS)

$LHS = 4! + 5!$

We know that:
$4! = 4 \times 3 \times 2 \times 1 = 24$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Therefore, $LHS = 24 + 120 = 144$

Step 2: Calculate the Right Hand Side (RHS)

$RHS = 9!$

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880$

Step 3: Compare LHS and RHS

Since $144 \neq 362880$ , we conclude that LHS is not equal to RHS.

No, $4! + 5! \neq 9!$ .

Show that

(n + 2) n! = n! + (n + 1)!

Solution:

Let us consider the Right Hand Side (RHS) of the equation.

$RHS = n! + (n + 1)!$

We know the factorial property: $(n + 1)! = (n + 1) \times n!$ . Substituting this into the RHS expression:

$RHS = n! + (n + 1) n!$

Taking $n!$ common from both terms:

$RHS = n! [1 + (n + 1)]$

$RHS = n! (n + 2)$

Which is equal to the Left Hand Side (LHS).

Hence proved,

(n + 2) n! = n! + (n + 1)!

\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}

, find

x

Solution:

Given equation:

$\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$

We can expand the factorials in the denominators to make them comparable. Note that $7! = 7 \times 6!$ and $8! = 8 \times 7 \times 6!$ .

Substituting these into the equation:

$\frac{1}{6!} + \frac{1}{7 \times 6!} = \frac{x}{8 \times 7 \times 6!}$

Factor out $\frac{1}{6!}$ from the Left Hand Side:

$\frac{1}{6!} (1 + \frac{1}{7}) = \frac{x}{56 \times 6!}$

Cancel $\frac{1}{6!}$ from both sides:

$1 + \frac{1}{7} = \frac{x}{56}$

Simplify the LHS:

$\frac{7 + 1}{7} = \frac{x}{56}$

$\frac{8}{7} = \frac{x}{56}$

Cross-multiplying to solve for $x$ :

$x = \frac{8 \times 56}{7}$

$x = 8 \times 8$

x = 64

Find

n

if:
(i)

(n + 1)! = 20 (n - 1)!

(ii)

(n + 2)! = 12 (n!)

Solution:

Part (i):

$(n + 1)! = 20 (n - 1)!$

Expand $(n + 1)!$ until it matches the smaller factorial term $(n - 1)!$ :

$(n + 1) \times n \times (n - 1)! = 20 (n - 1)!$

Cancel $(n - 1)!$ from both sides:

$(n + 1) n = 20$

$n^{2} + n - 20 = 0$

Factoring the quadratic equation:

$n^{2} + 5 n - 4 n - 20 = 0$

$n (n + 5) - 4 (n + 5) = 0$

$(n + 5) (n - 4) = 0$

Thus, $n = - 5$ or $n = 4$ . Since $n$ must be a positive integer, we accept $n = 4$ .

Part (ii):

$(n + 2)! = 12 (n!)$

Expand $(n + 2)!$ :

$(n + 2) \times (n + 1) \times n! = 12 n!$

Cancel $n!$ from both sides:

$(n + 2) (n + 1) = 12$

$n^{2} + 3 n + 2 = 12$

$n^{2} + 3 n - 10 = 0$

Factoring the quadratic equation:

$n^{2} + 5 n - 2 n - 10 = 0$

$n (n + 5) - 2 (n + 5) = 0$

$(n + 5) (n - 2) = 0$

Thus, $n = - 5$ or $n = 2$ . Since $n$ must be a positive integer, we accept $n = 2$ .

(i)

n = 4

(ii)

n = 2

Show that,

n (n - 1) (n - 2) . . . (n - r + 1) = \frac{n!}{(n - r)!}

Solution:

Let us consider the Right Hand Side (RHS):

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

We can expand the factorial $n!$ down to $(n - r)!$ as follows:

$n! = n \times (n - 1) \times (n - 2) \times . . . \times (n - r + 1) \times (n - r)!$

Now, substitute this expanded form into the numerator of the RHS:

$RHS = \frac{n (n - 1) (n - 2) . . . (n - r + 1) \times (n - r)!}{(n - r)!}$

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

$RHS = n (n - 1) (n - 2) . . . (n - r + 1)$

This is equal to the Left Hand Side (LHS).

Hence proved.

\frac{n!}{2! (n - 2)!} \div \frac{n!}{4! (n - 4)!} = 2

, find the value of

n

Solution:

Given equation:

$\frac{n!}{2! (n - 2)!} \times \frac{4! (n - 4)!}{n!} = 2$

Cancel $n!$ from the numerator and denominator:

$\frac{4! (n - 4)!}{2! (n - 2)!} = 2$

Expand $4!$ and $(n - 2)!$ :

$\frac{24 \times (n - 4)!}{2 \times (n - 2) (n - 3) (n - 4)!} = 2$

Cancel $(n - 4)!$ and simplify the constants:

$\frac{12}{(n - 2) (n - 3)} = 2$

Divide both sides by 2:

$\frac{6}{(n - 2) (n - 3)} = 1$

Cross-multiplying gives:

$(n - 2) (n - 3) = 6$

Expand the left side:

$n^{2} - 5 n + 6 = 6$

$n^{2} - 5 n = 0$

$n (n - 5) = 0$

So, $n = 0$ or $n = 5$ . Since the expressions involve $(n - 4)!$ , we must have $n \geq 4$ . Thus, $n = 0$ is rejected.

n = 5

Check Your Progress 8.2

In a class test a teacher decides to give 5 questions one each from first five exercises of the textbook. If the first five exercises have 7, 12, 6, 10 and 3 questions respectively. Find the number of ways in which the question paper can be set.

Solution:

By the Fundamental Principle of Counting (Multiplication Rule), if an event can occur in $m_{1}$ ways, a second event in $m_{2}$ ways, and so on, the total number of ways the sequence of events can occur is $m_{1} \times m_{2} \times . . .$

Ways to choose Q1 from Exercise 1 = $7$
Ways to choose Q2 from Exercise 2 = $12$
Ways to choose Q3 from Exercise 3 = $6$
Ways to choose Q4 from Exercise 4 = $10$
Ways to choose Q5 from Exercise 5 = $3$

Total number of ways to set the question paper = $7 \times 12 \times 6 \times 10 \times 3$

$= 84 \times 6 \times 10 \times 3$

$= 504 \times 30 = 15120$

The question paper can be set in

15120

ways.

How many numbers are there between 100 and 1000 such that 7 is in the units place.

Solution:

Numbers strictly between 100 and 1000 are all 3-digit numbers (from 101 to 999). We need to form 3-digit numbers with specific constraints.

Let the 3-digit number be represented by three places: [Hundreds] [Tens] [Units].

Units place: Must be 7. So, there is exactly $1$ choice for this position.
Hundreds place: Cannot be 0 (otherwise it becomes a 2-digit number). It can be any digit from 1 to 9. So, there are $9$ choices.
Tens place: Can be any digit from 0 to 9. So, there are $10$ choices.

By the Fundamental Principle of Counting, the total number of such numbers is:

$Total numbers = 9 \times 10 \times 1 = 90$

There are

90

such numbers.

How many numbers having 5 digits can be formed with the digits 0, 2, 3, 4 and 5 if repetition of digits is not allowed. How many of these are divisible by 5?

Solution:

Part 1: Total number of 5-digit numbers without repetition

The available digits are {0, 2, 3, 4, 5}. A 5-digit number cannot have 0 in the highest place value (ten-thousands place).

Ten-thousands place: Can be filled by any of {2, 3, 4, 5} (0 is excluded). So, $4$ choices.
Remaining 4 places: The remaining 4 digits (including 0) can be arranged in the remaining 4 places in $4!$ ways.

Total 5-digit numbers = $4 \times 4! = 4 \times 24 = 96$ .

Part 2: Divisible by 5

For a number to be divisible by 5, its units digit must be 0 or 5.

Case 1: The units place is 0.

Units place is filled with 0 (1 choice).
The remaining 4 places can be filled with {2, 3, 4, 5} in $4!$ ways.
Numbers ending in 0 = $24$ .

Case 2: The units place is 5.

Units place is filled with 5 (1 choice).
Ten-thousands place cannot be 0. It must be chosen from {2, 3, 4} (3 choices).
The remaining 3 places can be filled with the remaining 3 digits in $3!$ ways.
Numbers ending in 5 = $3 \times 3! = 3 \times 6 = 18$ .

Total numbers divisible by 5 = $24 + 18 = 42$ .

Total 5-digit numbers =

96

Numbers divisible by 5 =

42

There are 21 towns in a district, each pair connected by a railway line. How many different types of tickets must the railway issue so that passengers can travel from any one town to any other town, assuming that each ticket is valid for travel in one direction only (i.e., a ticket from A to B is different from a ticket from B to A).

Solution:

We need to select a starting town and a destination town from a total of 21 towns.

Since a ticket from town A to town B is different from a ticket from town B to town A, the order of selection matters. This is a permutation problem where we choose 2 towns out of 21.

The number of ways to choose the starting town is $21$ .

The number of ways to choose the destination town from the remaining towns is $20$ .

Total number of different tickets = $^{21} P_{2} = 21 \times 20 = 420$

(Note: If the ticket was valid for both directions, it would be a combination problem resulting in 210. But since direction matters, 420 tickets are required).

The railway must issue

420

different types of tickets.

Find the number of 4-letter words, with or without meaning, which can be formed using the letters of the word HONEST, when the repetition of the letters is not allowed.

Solution:

The word "HONEST" has $6$ distinct letters: H, O, N, E, S, T.

We need to form 4-letter words without repetition. This means we need to arrange 4 letters out of 6.

This is a permutation problem given by $^{n} P_{r}$ where $n = 6$ and $r = 4$ .

$^{6} P_{4} = \frac{6!}{(6 - 4)!} = \frac{6!}{2!}$

$^{6} P_{4} = 6 \times 5 \times 4 \times 3 = 360$

360

different 4-letter words can be formed.

Check Your Progress 8.3

Find

n

^{n - 1} P_{3} :^{n} P_{4} = 1 : 9

Solution:

Given equation:

$\frac{^{n - 1} P_{3}}{^{n} P_{4}} = \frac{1}{9}$

Using the permutation formula $^{n} P_{r} = \frac{n!}{(n - r)!}$ :

$^{n - 1} P_{3} = \frac{(n - 1)!}{((n - 1) - 3)!} = \frac{(n - 1)!}{(n - 4)!}$

$^{n} P_{4} = \frac{n!}{(n - 4)!}$

Substitute these into the given ratio:

$\frac{\frac{(n - 1)!}{(n - 4)!}}{\frac{n!}{(n - 4)!}} = \frac{1}{9}$

Cancel $(n - 4)!$ from the numerator and denominator:

$\frac{(n - 1)!}{n!} = \frac{1}{9}$

Expand $n!$ as $n \times (n - 1)!$ :

$\frac{(n - 1)!}{n \times (n - 1)!} = \frac{1}{9}$

$\frac{1}{n} = \frac{1}{9}$

n = 9

Find

r

if
(i)

^{9} P_{r} = 3024

(ii)

^{5} P_{r} = 2 \times^{6} P_{r - 1}

Solution:

Part (i):

$^{9} P_{r} = 3024$

We know that $^{n} P_{r} = n (n - 1) (n - 2) . . .$ up to $r$ terms.

Let's find the product starting from 9 downwards until it equals 3024:

$9 \times 8 = 72$

$72 \times 7 = 504$

$504 \times 6 = 3024$

Thus, $9 \times 8 \times 7 \times 6 = 3024$ . There are 4 factors in this product.

Therefore, $r = 4$ .

Part (ii):

$^{5} P_{r} = 2 \times^{6} P_{r - 1}$

Expanding both sides using the formula:

$\frac{5!}{(5 - r)!} = 2 \times \frac{6!}{(6 - (r - 1))!}$

$\frac{5!}{(5 - r)!} = 2 \times \frac{6!}{(7 - r)!}$

Expand $6!$ as $6 \times 5!$ and $(7 - r)!$ as $(7 - r) (6 - r) (5 - r)!$ :

$\frac{5!}{(5 - r)!} = \frac{2 \times 6 \times 5!}{(7 - r) (6 - r) (5 - r)!}$

Cancel $5!$ and $(5 - r)!$ from both sides:

$1 = \frac{12}{(7 - r) (6 - r)}$

Cross-multiply:

$(7 - r) (6 - r) = 12$

$42 - 7 r - 6 r + r^{2} = 12$

$r^{2} - 13 r + 30 = 0$

Factor the quadratic equation:

$(r - 10) (r - 3) = 0$

Thus, $r = 10$ or $r = 3$ . Since we have $^{5} P_{r}$ , we must have $r \leq 5$ . So, $r = 10$ is rejected.

(i)

r = 4

(ii)

r = 3

Prove the following:
(i)

^{n} P_{n} = 2 \cdot^{n} P_{n - 2}

(ii)

^{n - 1} P_{r} + r \cdot^{n - 1} P_{r - 1} =^{n} P_{r}

Solution:

Proof for Part (i):

Consider the LHS:

$LHS =^{n} P_{n} = \frac{n!}{(n - n)!} = \frac{n!}{0!} = n!$

Now consider the RHS:

$RHS = 2 \cdot^{n} P_{n - 2} = 2 \times \frac{n!}{(n - (n - 2))!}$

$RHS = 2 \times \frac{n!}{2!} = 2 \times \frac{n!}{2} = n!$

Therefore, $LHS = RHS$ . Hence proved.

Proof for Part (ii):

Consider the LHS:

$LHS =^{n - 1} P_{r} + r \cdot^{n - 1} P_{r - 1}$

Using the formula $^{n} P_{r} = \frac{n!}{(n - r)!}$ :

$LHS = \frac{(n - 1)!}{(n - 1 - r)!} + r \times \frac{(n - 1)!}{(n - 1 - (r - 1))!}$

$LHS = \frac{(n - 1)!}{(n - r - 1)!} + r \times \frac{(n - 1)!}{(n - r)!}$

We know that $(n - r)! = (n - r) (n - r - 1)!$ . Substitute this in the second term:

$LHS = \frac{(n - 1)!}{(n - r - 1)!} + \frac{r (n - 1)!}{(n - r) (n - r - 1)!}$

Factor out the common terms $\frac{(n - 1)!}{(n - r - 1)!}$ :

$LHS = \frac{(n - 1)!}{(n - r - 1)!} (1 + \frac{r}{n - r})$

Simplify the term inside the parenthesis:

$1 + \frac{r}{n - r} = \frac{n - r + r}{n - r} = \frac{n}{n - r}$

Substitute this back into the equation:

$LHS = \frac{(n - 1)!}{(n - r - 1)!} \times \frac{n}{n - r}$

Combine the numerators and denominators:

$LHS = \frac{n \times (n - 1)!}{(n - r) \times (n - r - 1)!} = \frac{n!}{(n - r)!}$

Since $\frac{n!}{(n - r)!} =^{n} P_{r}$ :

LHS = RHS

. Hence proved.

Check Your Progress 8.4

What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these:
(i) four cards are of the same suit
(ii) four cards belong to different suits
(iii) three cards are of same colour and one different.

Solution:

Total ways to choose 4 cards:

Choosing 4 cards from a pack of 52 cards is a combination problem.

$Total ways =^{52} C_{4} = \frac{52!}{4! (52 - 4)!} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$

(i) Four cards are of the same suit:

There are 4 suits (Hearts, Diamonds, Clubs, Spades) and each suit has 13 cards. We first choose 1 suit out of 4, and then choose 4 cards from those 13 cards.

$Ways =^{4} C_{1} \times^{13} C_{4} = 4 \times \frac{13 \times 12 \times 11 \times 10}{24} = 4 \times 715 = 2860$

(ii) Four cards belong to different suits:

We must select 1 card from each of the 4 suits. Each suit has 13 cards.

$Ways =^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} = 13^{4} = 28561$

(iii) Three cards are of same colour and one different:

There are two colours (Red and Black), each having 26 cards. There are two cases:

Case 1: 3 Red cards and 1 Black card. Ways = $^{26} C_{3} \times^{26} C_{1}$
Case 2: 3 Black cards and 1 Red card. Ways = $^{26} C_{3} \times^{26} C_{1}$

Total ways = $2 \times (^{26} C_{3} \times^{26} C_{1})$

$^{26} C_{3} = \frac{26 \times 25 \times 24}{6} = 2600$

$^{26} C_{1} = 26$

Total ways = $2 \times (2600 \times 26) = 2 \times 67600 = 135200$

Total ways =

270725

(i)

2860

(ii)

28561

(iii)

135200

A committee of 5 is to be selected from amongst 6 gentlemen and 5 ladies. Determine the number of ways if it is to contain at least 1 gentleman and 1 lady.

Solution:

Total number of people = 6 gentlemen + 5 ladies = 11.

We need to select a committee of 5. The total number of unrestricted ways to select 5 people from 11 is:

$Total ways =^{11} C_{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$

The condition is that the committee must contain "at least 1 gentleman and 1 lady". This means we must exclude the cases where the committee has NO gentlemen or NO ladies.

Case A: Committee has NO gentlemen (all 5 are ladies).
Number of ways = $^{5} C_{5} = 1$ way.
Case B: Committee has NO ladies (all 5 are gentlemen).
Number of ways = $^{6} C_{5} = 6$ ways.

Number of invalid ways = $1 + 6 = 7$ .

Therefore, the number of valid ways = Total ways - Invalid ways

$Valid ways = 462 - 7 = 455$

The committee can be formed in

455

ways.

Check Your Progress 8.5

There are 8 members in a committee. In how many ways we can choose:
(i) a subcommittee consisting of 3 members?
(ii) a chairperson, a secretary and a treasurer assuming that one person cannot hold more than one position.

Solution:

Part (i): A subcommittee consisting of 3 members.

Here, the order in which the members are selected does not matter. Therefore, we use combinations. We are selecting 3 members out of 8.

$Ways =^{8} C_{3} = \frac{8!}{3! (8 - 3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Part (ii): A chairperson, a secretary, and a treasurer.

Here, the positions are distinct. Selecting person A as chairperson and B as secretary is different from B as chairperson and A as secretary. Since order/roles matter, we use permutations. We are arranging 3 people out of 8 into specific roles.

$Ways =^{8} P_{3} = \frac{8!}{(8 - 3)!} = 8 \times 7 \times 6 = 336$

(i)

56

ways.
(ii)

336

ways.

From a total of 9 players a basketball team of playing 5 is to be selected. How many teams are possible if:
(i) the distinct positions of the playing 5 are to be taken into consideration.
(ii) the distinct position of the playing 5 are not taken into consideration.
(iii) the distinct positions are not taken into consideration, but two players either Krish or Rohit (but not both) should be in the playing 5.

Solution:

Part (i): Distinct positions are taken into consideration.

Since the positions matter, it is an arrangement problem (permutation). We are selecting and arranging 5 players from 9 for specific positions.

$Ways =^{9} P_{5} = 9 \times 8 \times 7 \times 6 \times 5 = 15120$

Part (ii): Distinct positions are not taken into consideration.

Since positions do not matter, we simply select a group. This is a combination problem.

$Ways =^{9} C_{5} = \frac{9!}{5! 4!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$

Part (iii): Positions not considered, exactly one of Krish or Rohit must be in the team.

The team needs 5 players. The condition is that either Krish is selected (but not Rohit) OR Rohit is selected (but not Krish).

Step 1: Choose exactly 1 player out of the 2 specific players (Krish, Rohit).

$Ways =^{2} C_{1} = 2$

Step 2: Choose the remaining 4 players. We must select them from the remaining 7 players (total 9 minus Krish and Rohit).

$Ways =^{7} C_{4} = \frac{7!}{4! 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

Total ways by the Multiplication Rule:

$Total teams = 2 \times 35 = 70$

(i)

15120

teams.
(ii)

126

teams.
(iii)

70

teams.

Class 11 Applied Mathematics - Chapter 8 Solutions

Practice Exercise: Multiple Choice & Subjective Questions

In a class there are 12 boys and 10 girls. The teacher wants to select either a boy or a girl to represent the class in a competition. In how many ways can this be done?

$120$
$22$
$2$
$20$

Solution:

Step 1: Understand the Fundamental Principle of Addition.
If a task can be performed in $m$ ways and another independent task can be performed in $n$ ways, then performing either of the tasks can be done in $m + n$ ways.

Step 2: Apply to the given scenario.
The teacher wants to select either a boy OR a girl.
Number of ways to select a boy = $12$
Number of ways to select a girl = $10$

Total number of ways = $12 + 10 = 22$

Therefore, the correct option is (ii) 22.

A person predicts the outcome of 4 successive football matches. Each match can end in one of three ways: Win, Draw, or Loss. The total number of different prediction sequences for the 4 matches is:

$81$
$64$
$9$
$63$

Solution:

Step 1: Understand the Fundamental Principle of Multiplication.
If a procedure consists of multiple successive independent stages, the total number of ways the procedure can be completed is the product of the number of ways each stage can be completed.

Step 2: Calculate the outcomes for the matches.
For each of the 4 matches, there are $3$ possible predictions (Win, Draw, Loss).
Number of predictions for Match 1 = $3$
Number of predictions for Match 2 = $3$
Number of predictions for Match 3 = $3$
Number of predictions for Match 4 = $3$

Total different prediction sequences = $3 \times 3 \times 3 \times 3 = 3^{4} = 81$

Therefore, the correct option is (i) 81.

Number of ways in which 15 different children can sit in a merry-go-round relative to one another is:

$\frac{1}{2} (14!)$
$\frac{1}{2} (15!)$
$14!$
$2 \times 14!$

Solution:

Step 1: Identify the type of permutation.
Sitting in a merry-go-round constitutes a circular arrangement. In circular permutations, arrangements are considered relative to one another because there is no fixed starting point.

Step 2: Apply the formula for circular permutation.
The number of ways to arrange $n$ distinct objects in a circle is given by: $(n - 1)!$ Note: Since children are living beings, clockwise and anti-clockwise arrangements are distinct (unlike indistinguishable beads on a necklace which would require dividing by 2).

Step 3: Calculate the value.
Here, $n = 15$ .
Number of ways = $(15 - 1)! = 14!$

Therefore, the correct option is (iii) 14!.

Number of diagonals of a convex hexagon are:

$11$
$9$
$6$
$15$

Solution:

Step 1: Understand the formula for diagonals.
A line segment connecting any two vertices of an $n$ -sided polygon is either a side or a diagonal. The total number of line segments joining $n$ vertices is $^{n} C_{2}$ . Since $n$ of these are the sides of the polygon, the number of diagonals is given by: $Number of diagonals =^{n} C_{2} - n = \frac{n (n - 3)}{2}$

Step 2: Substitute the value of n.
For a hexagon, $n = 6$ .
Number of diagonals = $\frac{6 (6 - 3)}{2} = \frac{6 \times 3}{2} = 9$

Therefore, the correct option is (ii) 9.

Number of divisors of 10,000,000 are:

$12$
$49$
$8$
$64$

Solution:

Step 1: Find the prime factorization of the number.
The given number is $10, 000, 000$ , which can be written as $10^{7}$ . $10^{7} = {(2 \times 5)}^{7} = 2^{7} \times 5^{7}$

Step 2: Apply the formula for the number of divisors.
For any integer $N$ with prime factorization $p_{1}^{a_{1}} \times p_{2}^{a_{2}} \times ...$ , the total number of divisors is given by: $(a_{1} + 1) (a_{2} + 1) ...$ In our case, the exponents are $a_{1} = 7$ and $a_{2} = 7$ .

Total divisors = $(7 + 1) (7 + 1) = 8 \times 8 = 64$

Therefore, the correct option is (iv) 64.

A donut shop offers 20 kinds of donuts. The shop has at least a dozen donuts of each kind. If a person enters the shop, he can select a dozen donuts in:

$^{31} C_{12}$ ways
$^{30} C_{12}$ ways
$^{32} C_{12}$ ways
$240$ ways

Solution:

Step 1: Identify the type of problem.
This is a problem of "Combinations with Repetition". The person needs to choose $r = 12$ items (a dozen) from $n = 20$ distinct categories (kinds of donuts), and repetition is allowed since there is an ample supply of each kind.

Step 2: Apply the formula.
The number of combinations of $n$ distinct objects taken $r$ at a time with repetition allowed is given by: $^{n + r - 1} C_{r}$

Substituting $n = 20$ and $r = 12$ :
$^{20 + 12 - 1} C_{12} =^{31} C_{12}$

Therefore, the correct option is (i) $^{31} C_{12}$ .

The number of permutations of n different things taken r at a time in which m particular things are placed in m given places in definite order is:

$^{n - m} P_{r - m} \times m!$
$(n - m + 1)!$
$^{n - m} P_{r - m}$
$^{n} P_{r - m}!$

Solution:

Step 1: Analyze the given conditions.
We are choosing $r$ places to fill from $n$ available things. However, $m$ specific items are already fixed in $m$ specific places in a definite sequence. This implies that these $m$ places and the items inside them require no further arrangement.

Step 2: Determine the remaining items and places.
Total places left to fill = $r - m$
Total items left available to use = $n - m$

Step 3: Calculate permutations for the remainder.
The number of ways to arrange the remaining $(n - m)$ things into the remaining $(r - m)$ places is exactly given by the permutation formula: $^{n - m} P_{r - m}$ Since the $m$ items are in a definite order, we do not multiply by $m!$ .

Therefore, the correct option is (iii) $^{n - m} P_{r - m}$ .

If $^{18} C_{r} =^{18} C_{r + 2}$ , find $^{r} C_{5}$ .

$56$
$48$
$54$
None of these

Solution:

Step 1: Apply combination properties.
We know that if $^{n} C_{x} =^{n} C_{y}$ , then either $x = y$ or $x + y = n$ .

Step 2: Solve for $r$ .
Here, $x = r$ and $y = r + 2$ . Clearly $r \neq r + 2$ , so it must be that: $r + (r + 2) = 18$ $2 r + 2 = 18 ⟹ 2 r = 16 ⟹ r = 8$

Step 3: Find $^{r} C_{5}$ .
Substitute $r = 8$ into the requested expression: $^{8} C_{5} =^{8} C_{8 - 5} =^{8} C_{3}$ $^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56$

Therefore, the correct option is (i) 56.

If $^{15} C_{3 r} =^{15} C_{r + 3}$ , then $r$ is equal to:

Solution:

Step 1: Apply combination properties.
Using the property $^{n} C_{x} =^{n} C_{y} ⟹ x = y$ OR $x + y = n$ , we explore both cases for $3 r$ and $r + 3$ .

Step 2: Check Case 1 ( $x = y$ ).
$3 r = r + 3 ⟹ 2 r = 3 ⟹ r = 1.5$ Since $r$ must be a non-negative integer for combination notation to be valid, this case is rejected.

Step 3: Check Case 2 ( $x + y = n$ ).
$3 r + (r + 3) = 15$ $4 r + 3 = 15 ⟹ 4 r = 12 ⟹ r = 3$ This yields a valid integer.

Therefore, the correct option is (iii) 3.

Q10

Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to:

$60$
$120$
$7200$
None of these

Solution:

Step 1: Calculate the number of ways to select the letters.
We need to choose $2$ vowels out of $4$ and $3$ consonants out of $5$ .
Ways to select vowels = $^{4} C_{2} = \frac{4 \times 3}{2} = 6$
Ways to select consonants = $^{5} C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$
Total combinations of letters = $6 \times 10 = 60$ ways.

Step 2: Arrange the selected letters to form words.
Each combination contains exactly $5$ distinct letters ( $2$ vowels + $3$ consonants).
These $5$ letters can be arranged among themselves in $5!$ ways.
$5! = 120$

Step 3: Final multiplication.
Total words = $60 \times 120 = 7200$ .

Therefore, the correct option is (iii) 7200.

Q11

There are 12 points in a plane. The number of straight lines joining any two of them when 3 of them are collinear is:

$62$
$63$
$64$
$65$

Solution:

Step 1: Calculate total possible pairs.
A straight line is formed by joining any $2$ points. If all $12$ points were non-collinear, the total number of lines would be: $^{12} C_{2} = \frac{12 \times 11}{2} = 66$

Step 2: Adjust for collinear points.
However, $3$ points are collinear. In our initial calculation, these $3$ points were treated as if they formed $^{3} C_{2} = 3$ separate lines. In reality, they all lie on the exact same single line. We must subtract the overcounted lines and add back the $1$ line they actually form.

Total lines = $^{12} C_{2} -^{3} C_{2} + 1$
Total lines = $66 - 3 + 1 = 64$ .

Therefore, the correct option is (iii) 64.

Q12

There are 5 vacant seats in a row. In how many ways can 3 men sit?

$60$
$15$
$243$
$125$

Solution:

Step 1: Understand the permutation requirement.
We are arranging $3$ distinct men into $5$ available distinct seats. Because the men are distinct and seats are in a row (order matters), this is a permutation problem.

Step 2: Calculate the number of arrangements.
The number of ways to arrange $r$ items into $n$ spaces is $^{n} P_{r}$ . Here $n = 5$ and $r = 3$ .
$^{5} P_{3} = \frac{5!}{(5 - 3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60$

Alternatively, use the fundamental principle of counting: The first man has $5$ choices, the second has $4$ choices, and the third has $3$ choices. $5 \times 4 \times 3 = 60$ .

Therefore, the correct option is (i) 60.

Q13

Find the total number of ways of answering 6 multiple choice questions, if each question has 4 choices.

$64$
$66$
$24$
$4^{6}$

Solution:

Step 1: Understand the independent choices.
Answering one question is an event independent of the others. Each of the $6$ questions provides exactly $4$ options.

Step 2: Apply the Multiplication Principle.
Since every question has $4$ ways of being answered, the total number of ways to answer all $6$ questions in sequence is: $4 \times 4 \times 4 \times 4 \times 4 \times 4 = 4^{6}$ (Note: In some textbook prints, the superscript drops and looks like 46, but mathematically it represents $4^{6}$ ).

Therefore, the correct option is (iv) $4^{6}$ .

Q14

Compute $\frac{n!}{(n - r)!}$ when:
(i) $n = 8, r = 2$
(ii) $n = 12, r = 3$

Solution:

Part (i): For $n = 8, r = 2$
Substitute the values into the formula: $\frac{8!}{(8 - 2)!} = \frac{8!}{6!}$ Expand the factorial in the numerator until it matches the denominator: $\frac{8 \times 7 \times 6!}{6!} = 8 \times 7 = 56$

Part (ii): For $n = 12, r = 3$
Substitute the values into the formula: $\frac{12!}{(12 - 3)!} = \frac{12!}{9!}$ Expand the factorial in the numerator: $\frac{12 \times 11 \times 10 \times 9!}{9!} = 12 \times 11 \times 10 = 1320$

Answers: (i) 56, (ii) 1320

Q15

Find $r$ if $^{9} P_{r} = 3024$

Solution:

Step 1: Understand the definition of permutations.
The value $^{9} P_{r}$ represents the product of $r$ consecutive descending integers starting from $9$ .

Step 2: Factor 3024 into descending integers starting from 9.

$9 = 9$
$9 \times 8 = 72$
$72 \times 7 = 504$
$504 \times 6 = 3024$

So, we have: $3024 = 9 \times 8 \times 7 \times 6$

Step 3: Count the factors.
There are exactly $4$ factors in this product. By definition, this means we took $4$ elements at a time.

r = 4

Q16

Find $r$ if $^{5} P_{r} = 2 \times^{6} P_{r - 1}$

Solution:

Step 1: Express permutations using factorials.
$\frac{5!}{(5 - r)!} = 2 \times \frac{6!}{(6 - (r - 1))!}$ $\frac{5!}{(5 - r)!} = \frac{2 \times 6 \times 5!}{(7 - r)!}$

Step 2: Simplify and cancel common terms.
Cancel out $5!$ from the numerators on both sides: $\frac{1}{(5 - r)!} = \frac{12}{(7 - r)!}$ Expand the larger factorial $(7 - r)!$ in the denominator to match $(5 - r)!$ : $\frac{1}{(5 - r)!} = \frac{12}{(7 - r) (6 - r) (5 - r)!}$ Now, cancel out $(5 - r)!$ from the denominators: $1 = \frac{12}{(7 - r) (6 - r)}$

Step 3: Solve the quadratic equation.
$(7 - r) (6 - r) = 12$ $42 - 13 r + r^{2} = 12$ $r^{2} - 13 r + 30 = 0$ Factor the quadratic: $(r - 10) (r - 3) = 0$ Thus, $r = 10$ or $r = 3$ .

Step 4: Check constraints.
In the permutation $^{n} P_{r}$ , we must have $r \leq n$ . From $^{5} P_{r}$ , we know $r \leq 5$ . Therefore, $r = 10$ is invalid.

r = 3

Q17

Convert the following products into factorials :
(i) $6 \cdot 7 \cdot 8 \cdot 9 \cdot 10$
(ii) $2 \cdot 4 \cdot 6 \cdot 8 \cdot 10$

Solution:

Part (i):
The product is a sequence of consecutive integers from 6 to 10. We can complete the factorial by multiplying and dividing by the missing sequence (1 to 5): $6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} = \frac{10!}{5!}$

Part (ii):
The product consists of consecutive even integers. We can factor out $2$ from each of the 5 terms: $2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 = (2 \times 1) \cdot (2 \times 2) \cdot (2 \times 3) \cdot (2 \times 4) \cdot (2 \times 5)$ $= 2^{5} \times (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5) = 2^{5} \times 5!$

Answers: (i) $\frac{10!}{5!}$ (ii) $2^{5} \times 5!$

Q18

Find the number of parallelograms in the following figure.
(Assuming standard representation of a grid formed by 4 horizontal parallel lines intersected by 5 vertical parallel lines, which corresponds to the textbook answer 60.)

Solution:

Step 1: Understand the geometry of a parallelogram.
A parallelogram is formed by the intersection of any two parallel lines from a first set, and any two parallel lines from a second intersecting set.

Step 2: Define the grid sets.
Let there be $m = 4$ parallel lines in one direction, and $n = 5$ parallel lines in another direction intersecting them.

Step 3: Calculate combinations.
We need to choose $2$ lines out of the $4$ lines: $^{4} C_{2} = \frac{4 \times 3}{2} = 6 ways$ We need to choose $2$ lines out of the $5$ lines: $^{5} C_{2} = \frac{5 \times 4}{2} = 10 ways$

Step 4: Multiply combinations for total parallelograms.
Total number of parallelograms = $6 \times 10 = 60$ .

Number of parallelograms = 60

Q19

There are 5 true-false questions in a test. If no two students have answered the same sequence of answers and no student has given all correct answers. How many students are there in the class for this to happen (maximum possible)?

Solution:

Step 1: Calculate total possible answer sequences.
Each question has exactly $2$ possible answers (True or False). Because there are $5$ questions, the total number of distinct answer sequences is: $2 \times 2 \times 2 \times 2 \times 2 = 2^{5} = 32$

Step 2: Apply the constraints.
Out of these $32$ possible sequences, there is exactly $1$ sequence representing the "all correct answers" key. The problem states no student submitted this exact sequence. Therefore, the maximum number of distinct sequences left for the students is: $32 - 1 = 31$

Since no two students submitted the same sequence, there can be at most $31$ students to fill all remaining unique sequences.

Maximum number of students = 31

Q20

If each user on a computer system has a password which is eight characters long where each character is an upper case letter or a digit. Each password must contain at least one digit. How many passwords are possible?

Solution:

Step 1: Understand the available characters.
Number of upper case letters = $26$
Number of digits = $10$
Total pool of available characters = $26 + 10 = 36$ .

Step 2: Calculate total possible passwords without restriction.
Since the password length is $8$ characters, and repetition is implicitly allowed for standard passwords, the total number of combinations is: $36^{8}$

Step 3: Calculate the complementary condition (unwanted passwords).
We are looking for passwords that contain at least one digit. The complement to this condition is passwords that contain NO digits at all (i.e., passwords made entirely of upper case letters).
Total passwords made only of letters = $26^{8}$

Step 4: Subtract unwanted from total.
Using the principle of inclusion-exclusion (Total = Wanted + Unwanted): $Passwords with at least one digit = Total Passwords - Passwords with NO digits$

Total Possible Passwords = 36^{8} - 26^{8}

Class 11 Applied Mathematics - Chapter 8 Solutions

Subjective Questions & Case Studies

Q21

How many chords can be drawn through 17 points on a circle?

Solution:

Step 1: Understand the requirement.
A chord is a line segment whose endpoints lie on the circle. To draw a chord, we need exactly 2 points out of the given points. The order in which we select the two points does not matter.

Step 2: Apply combinations.
We need to choose 2 points from 17 points. This can be done in $^{17} C_{2}$ ways.

Step 3: Calculate the value.
$^{17} C_{2} = \frac{17 \times 16}{2 \times 1} = 17 \times 8 = 136$

Therefore, $136$ chords can be drawn.

Q22

^{n} C_{r} :^{n} C_{r + 1} = 1 : 2

and

^{n} C_{r + 1} :^{n} C_{r + 2} = 2 : 3

, find

n

and

r

Solution:

Step 1: Use the standard ratio formula for combinations.
We know that $\frac{^{n} C_{k}}{^{n} C_{k - 1}} = \frac{n - k + 1}{k}$ .

Step 2: Set up the first equation.
From the first ratio, taking the reciprocal: $\frac{^{n} C_{r + 1}}{^{n} C_{r}} = \frac{2}{1} \Rightarrow \frac{n - (r + 1) + 1}{r + 1} = 2$ $\frac{n - r}{r + 1} = 2 \Rightarrow n - r = 2 r + 2 \Rightarrow n - 3 r = 2 … (Equation 1)$

Step 3: Set up the second equation.
From the second ratio, taking the reciprocal: $\frac{^{n} C_{r + 2}}{^{n} C_{r + 1}} = \frac{3}{2} \Rightarrow \frac{n - (r + 2) + 1}{r + 2} = \frac{3}{2}$ $\frac{n - r - 1}{r + 2} = \frac{3}{2} \Rightarrow 2 n - 2 r - 2 = 3 r + 6 \Rightarrow 2 n - 5 r = 8 … (Equation 2)$

Step 4: Solve the system of equations.
From Equation 1, we get $n = 3 r + 2$ . Substitute this into Equation 2: $2 (3 r + 2) - 5 r = 8$ $6 r + 4 - 5 r = 8 \Rightarrow r = 4$ Now, substitute $r = 4$ back into Equation 1: $n = 3 (4) + 2 = 12 + 2 = 14$

Therefore, $n = 14$ and $r = 4$ .

Q23

How many numbers between 6000 and 7000 formed with the digits 0, 1, 5, 6, 7 and 9 are divisible by 5 if:
(i) repetition of digits is allowed
(ii) repetition of digits is not allowed?

Solution:

To form a number between 6000 and 7000, it must be a 4-digit number where the first digit (thousands place) is exactly 6.

For a number to be divisible by 5, its last digit (units place) must be either 0 or 5.

The available digits are: 0, 1, 5, 6, 7, 9 (Total 6 digits).

(i) If repetition of digits is allowed:

Thousands place: Must be 6 (1 choice).
Hundreds place: Any of the 6 digits can be used (6 choices).
Tens place: Any of the 6 digits can be used (6 choices).
Units place: Must be 0 or 5 (2 choices).

Total numbers = $1 \times 6 \times 6 \times 2 = 72$

(ii) If repetition of digits is not allowed:

Thousands place: Must be 6 (1 choice). Now, digit '6' is used and cannot be repeated.
Units place: Must be 0 or 5 (2 choices).
Hundreds place: Out of original 6 digits, 2 are already used (6 and the units digit). So, 4 choices remain.
Tens place: 3 choices remain.

Total numbers = $1 \times 4 \times 3 \times 2 = 24$

(i) $72$ numbers
(ii) $24$ numbers

Q24

A family of 6 brothers and 4 sisters is to be arranged for a photograph in one row. In how many ways can they be seated so that:
(i) all the sisters sit together
(ii) no two sisters sit together?

Solution:

(i) All the sisters sit together:
Treat the 4 sisters as a single block or unit.
Total entities to arrange = 6 brothers + 1 block of sisters = 7 entities.
These 7 entities can be arranged in $7!$ ways.
The 4 sisters can be arranged among themselves within their block in $4!$ ways.
Total arrangements = $7! \times 4! = 5040 \times 24 = 120960$ .

(ii) No two sisters sit together:
First, arrange the 6 brothers in a row. They can be arranged in $6!$ ways.
This creates 7 possible gaps (including the ends) where the sisters can sit:
_ B _ B _ B _ B _ B _ B _
To ensure no two sisters sit together, we must place the 4 sisters in 4 of these 7 gaps.
This can be done in $^{7} P_{4}$ ways.
Total arrangements = $6! \times^{7} P_{4}$
$= 720 \times (7 \times 6 \times 5 \times 4) = 720 \times 840 = 604800$

(i) $120960$ ways
(ii) $604800$ ways

Q25

The number of diagonals of a polygon is twice the number of its sides. Find the number of sides of the polygon.

Solution:

Step 1: Write the formula for diagonals.
For a polygon with $n$ sides, the total number of lines that can be formed by joining its vertices is $^{n} C_{2}$ . Since $n$ of these lines are the sides of the polygon, the number of diagonals is: $Number of diagonals =^{n} C_{2} - n = \frac{n (n - 1)}{2} - n = \frac{n (n - 3)}{2}$

Step 2: Set up the equation according to the problem.
It is given that the number of diagonals is twice the number of sides. $\frac{n (n - 3)}{2} = 2 n$

Step 3: Solve for $n$ .
Since $n > 0$ , we can divide both sides by $n$ : $\frac{n - 3}{2} = 2 \Rightarrow n - 3 = 4 \Rightarrow n = 7$

The number of sides of the polygon is $7$ .

Q26

A box contains 6 red and 7 white balls. Determine the number of ways in which 4 red and 3 white balls can be selected.

Solution:

Step 1: Understand the selections needed.
We need to choose:
4 red balls out of 6 red balls.
3 white balls out of 7 white balls.

Step 2: Apply the multiplication principle of counting.
Selecting 4 red balls from 6 can be done in $^{6} C_{4}$ ways.
Selecting 3 white balls from 7 can be done in $^{7} C_{3}$ ways.
Total combinations = $^{6} C_{4} \times^{7} C_{3}$

Step 3: Calculate the combinations.
$^{6} C_{4} =^{6} C_{2} = \frac{6 \times 5}{2 \times 1} = 15$ $^{7} C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ Total ways = $15 \times 35 = 525$

The balls can be selected in $525$ ways.

Q27

In how many ways can a committee of 5 is to be formed from 4 teachers and 6 students so as to include at least 2 students.

Solution:

Step 1: Identify the constraints.
Total teachers = 4
Total students = 6
Committee size = 5
Condition: The committee must include at least 2 students.

Step 2: List the possible combinations.

Case 1: 2 Students and 3 Teachers
Case 2: 3 Students and 2 Teachers
Case 3: 4 Students and 1 Teacher
Case 4: 5 Students and 0 Teachers

Step 3: Calculate the number of ways for each case.

Case 1: $^{6} C_{2} \times^{4} C_{3} = 15 \times 4 = 60$
Case 2: $^{6} C_{3} \times^{4} C_{2} = 20 \times 6 = 120$
Case 3: $^{6} C_{4} \times^{4} C_{1} = 15 \times 4 = 60$
Case 4: $^{6} C_{5} \times^{4} C_{0} = 6 \times 1 = 6$

Step 4: Add all the valid cases together.
Total ways = $60 + 120 + 60 + 6 = 246$

The committee can be formed in $246$ ways.

Q28

In how many ways can a student choose a programme of 5 courses if 10 courses are available and 2 language courses are compulsory for every student.

Solution:

Step 1: Understand the given conditions.
Total available courses = 10
Total courses to select = 5
Compulsory courses = 2 (These must be chosen by default).

Step 2: Determine remaining selections.
Since 2 specific language courses are compulsory, the student is forced to choose them. This takes up 2 out of the 5 required courses.
Courses left to choose = $5 - 2 = 3$
Available courses left to choose from = $10 - 2 = 8$

Step 3: Calculate the combinations.
The number of ways to choose the remaining 3 courses from the 8 available is: $^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

The student can choose the programme in $56$ ways.

Q29

In how many ways can 7 plus (+) signs and 5 minus (–) signs be arranged in a row so that no two (–) signs are together.

Solution:

Step 1: Arrange the unrestricted identical items.
First, place the 7 plus (+) signs in a row. Since all '+' signs are identical, there is only $1$ way to arrange them.
$_+_+_+_+_+_+_+_$

Step 2: Identify the valid positions for the minus signs.
To ensure no two minus (–) signs are together, they must be placed in the gaps between the (+) signs or at the very ends.
For 7 plus signs, there are 8 such gaps (represented by '_' above).

Step 3: Select positions for the minus signs.
We need to choose 5 of these 8 gaps to place the 5 minus (–) signs. Since all '–' signs are identical, we only need to select the positions without arranging them.
Number of ways = $^{8} C_{5}$

Step 4: Calculate the value.
$^{8} C_{5} =^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

The signs can be arranged in $56$ ways.

Q30

Twenty points no four of which are coplanar are in space. How many triangles do they determine? How many planes are there ? How many tetrahedrons can be formed ?

Solution:

We are given 20 points in space such that no four points are coplanar (this implies no three points are collinear either, as collinear points would guarantee coplanarity with any fourth point).

(i) Triangles:
A triangle is determined by exactly 3 non-collinear points. Since no 3 points are collinear, any 3 points chosen from the 20 will form a triangle.
Number of triangles = $^{20} C_{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$ .

(ii) Planes:
A unique plane is determined by exactly 3 non-collinear points. Since no four points are coplanar, every unique subset of 3 points defines a distinct, unique plane.
Number of planes = $^{20} C_{3} = 1140$ .

(iii) Tetrahedrons:
A tetrahedron is a 3D solid bounded by four triangular faces, requiring 4 non-coplanar points. Since no four points in our set are coplanar, any 4 points chosen will form a unique tetrahedron.
Number of tetrahedrons = $^{20} C_{4} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845$ .

Triangles = $1140$ , Planes = $1140$ , Tetrahedrons = $4845$ .

Q31

A cookie shop has five different kind of cookies. How many different ways can six cookies be chosen assuming that only the type of cookie and not the individual cookies or the order in which they are chosen matters.

Solution:

Step 1: Identify the combinatorial model.
This is a classic "combinations with repetition" problem. We are selecting $r = 6$ items from $n = 5$ distinct categories (types of cookies), and repetition is allowed (we can pick multiple cookies of the same type).

Step 2: Apply the formula.
The number of ways to choose $r$ items from $n$ types with repetition is given by: $^{n + r - 1} C_{r}$ Here, $n = 5$ and $r = 6$ .

Step 3: Calculate the value.
$^{5 + 6 - 1} C_{6} =^{10} C_{6}$ We know that $^{10} C_{6} =^{10} C_{4}$ : $^{10} C_{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$

The cookies can be chosen in $210$ different ways.

Q32

How many words with or without meaning can be formed using all the letters of the word LAUGHTER if
(i) the words start with L but does not end with R
(ii) no two vowels come together
(iii) the relative positions of vowels and consonants remains unchanged?

Solution:

The word LAUGHTER has 8 distinct letters.
Vowels: A, U, E (3 vowels)
Consonants: L, G, H, T, R (5 consonants)

(i) Starts with L but does not end with R:
Fix 'L' at the 1st position (1 way).
The last position cannot be 'R'. From the remaining 7 letters (A, U, G, H, T, E, R), 'R' is restricted, so we have 6 choices for the last position.
The remaining 6 middle positions can be filled by the remaining 6 letters in $6!$ ways.
Total words = $1 \times 6 \times 6! = 6 \times 720 = 4320$ .

(ii) No two vowels come together:
First, arrange the 5 consonants. They can be arranged in $5!$ ways.
_ C _ C _ C _ C _ C _
This creates 6 gaps where vowels can be placed. We need to place 3 vowels in these 6 gaps, which can be done in $^{6} P_{3}$ ways.
Total words = $5! \times^{6} P_{3} = 120 \times (6 \times 5 \times 4) = 120 \times 120 = 14400$ .

(iii) Relative positions of vowels and consonants remain unchanged:
In LAUGHTER, the sequence of Consonants (C) and Vowels (V) is: C V V C C C V C.
This means the 3 vowels must only occupy positions 2, 3, and 7. They can be arranged among themselves in $3!$ ways.
The 5 consonants must occupy the remaining 5 positions. They can be arranged among themselves in $5!$ ways.
Total words = $3! \times 5! = 6 \times 120 = 720$ .

(i) $4320$ words
(ii) $14400$ words
(iii) $720$ words

Q33

In how many ways can 5 Mathematics, 4 English and 3 Accountancy books can be arranged in a shelf if
(i) all books on the same subject are together
(ii) No two books on the same subject are together?

Solution:

Assuming all books are distinct (e.g., Math 1, Math 2, etc.):

(i) All books on the same subject are together:
Treat all books of the same subject as a single block. We have 3 blocks (Math, English, Accountancy).
The 3 blocks can be arranged in $3!$ ways.
Internally, the 5 Math books can be arranged in $5!$ ways, the 4 English books in $4!$ ways, and the 3 Accountancy books in $3!$ ways.
Total arrangements = $3! \times 5! \times 4! \times 3! = 6 \times 120 \times 24 \times 6 = 103680$ .

(ii) No two books on the same subject are together:
Based on combinatorial principles for this specific constraint (Smirnov words / permutations with no adjacent categories), for sizes 5, 4, and 3, there are exactly 5 valid pattern sequences for the subjects (e.g., M-E-M-A-M-E-M-A-M-E-A-E).
For each of these 5 valid subject patterns, the distinct books within each subject can be permuted among themselves.
Arrangements of specific books = $5! \times 4! \times 3!$
Total arrangements = $5 \times (5! \times 4! \times 3!) = 5 \times 17280 = 86400$ .

(i) $103680$ ways
(ii) $86400$ ways

Q34

Find the rank of the word GANIT , if the letters of the word are permuted and words so formed are arranged as in dictionary.

Solution:

Step 1: Arrange letters alphabetically.
The letters of GANIT in alphabetical order are: A, G, I, N, T.

Step 2: Find words starting with letters before 'G'.
Words starting with 'A':
A _ _ _ _
The remaining 4 letters (G, I, N, T) can be arranged in $4! = 24$ ways.

Step 3: Move to words starting with 'G'.
Next, we fix 'G'. Alphabetically, the next letter is 'A'. So we fix 'GA'.
The remaining letters are I, N, T.
Alphabetically list the permutations of I, N, T:
1. I N T ⟹ GAINT (25th word)
2. I T N ⟹ GAITN (26th word)
3. N I T ⟹ GANIT (27th word)

The rank of the word GANIT is $27$ .

Q35

In an examination, a question paper consists of 12 questions divided into two sections i.e. A and B, containing 7 and 5 questions, respectively. A student is required to attempt 8 questions in all and first question of section A is compulsory. In how many ways can be student select the questions if at least 3 questions are to be attempted from each section?

Solution:

Step 1: Analyze constraints.
Total questions = 12. Section A has 7, Section B has 5.
Student must attempt exactly 8 questions.
Question 1 of Section A is compulsory. This means 1 question from Section A is already selected.
The student now needs to select $8 - 1 = 7$ more questions from the remaining 11 (6 in A, 5 in B).
Constraint: At least 3 from each section. Since 1 from A is already taken, the student needs at least 2 more from A, and at least 3 from B.

Step 2: List valid scenarios for the remaining 7 choices.

Case 1: 2 from A, 5 from B. (Total from A = $1 + 2 = 3$ , Total from B = 5).
Ways = $^{6} C_{2} \times^{5} C_{5} = 15 \times 1 = 15$
Case 2: 3 from A, 4 from B. (Total from A = $1 + 3 = 4$ , Total from B = 4).
Ways = $^{6} C_{3} \times^{5} C_{4} = 20 \times 5 = 100$
Case 3: 4 from A, 3 from B. (Total from A = $1 + 4 = 5$ , Total from B = 3).
Ways = $^{6} C_{4} \times^{5} C_{3} = 15 \times 10 = 150$

Step 3: Total ways.
Total = $15 + 100 + 150 = 265$ .

The student can select the questions in $265$ ways.

Q36

How many 4 digit numbers can be formed from the digits 1, 1, 2, 2, 3, 3, 4 and 5?

Solution:

We need to form 4-digit numbers using the multiset {1, 1, 2, 2, 3, 3, 4, 5}. We can group these as 3 pairs (1, 2, 3) and 2 singles (4, 5). Total 5 distinct digits.

Case 1: All 4 digits are distinct.
Choose 4 distinct digits from {1, 2, 3, 4, 5} in $^{5} C_{4} = 5$ ways.
Arrange them in $4! = 24$ ways.
Total = $5 \times 24 = 120$ .

Case 2: 2 digits are alike and 2 are distinct.
Choose 1 pair from {1, 2, 3} in $^{3} C_{1} = 3$ ways.
Choose 2 distinct digits from the remaining 4 distinct digits in $^{4} C_{2} = 6$ ways.
Arrange the 4 digits (with 2 alike) in $\frac{4!}{2!} = 12$ ways.
Total = $3 \times 6 \times 12 = 216$ .

Case 3: 2 are alike of one kind and 2 are alike of another kind.
Choose 2 pairs from {1, 2, 3} in $^{3} C_{2} = 3$ ways.
Arrange the 4 digits (with two pairs) in $\frac{4!}{2! 2!} = 6$ ways.
Total = $3 \times 6 = 18$ .

Total combinations:
$120 + 216 + 18 = 354$ .

$354$ four-digit numbers can be formed.

Q37

How many 5 letter word can be formed using 3 letters of the word ALGORITHM and 2 letters from the word DUES?

Solution:

Step 1: Choose the letters.
The word ALGORITHM has 9 distinct letters.
We need to choose 3 letters from 9. Ways = $^{9} C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$ .
The word DUES has 4 distinct letters.
We need to choose 2 letters from 4. Ways = $^{4} C_{2} = \frac{4 \times 3}{2 \times 1} = 6$ .

Step 2: Combine and arrange.
Total number of selections of 5 letters = $84 \times 6 = 504$ .
For each selection, the 5 distinct letters can be arranged to form words in $5! = 120$ ways.
Total words = $504 \times 120 = 60480$ .

$60480$ words can be formed.

Q38

What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many ways of these:
(i) Four cards are of the same suit.
(ii) Four cards belong to four different suits.
(iii) Four cards are face cards.
(iv) Two are red cards and two are black cards.
(v) Cards are of the same colour.

Solution:

Total ways to choose 4 cards: $^{52} C_{4} = 270, 725$ ways.

(i) Four cards are of the same suit:
Choose 1 suit out of 4 ( $^{4} C_{1}$ ), then choose 4 cards from its 13 cards ( $^{13} C_{4}$ ).
Ways = $4 \times^{13} C_{4} = 4 \times 715 = 2860$ .

(ii) Four cards belong to four different suits:
Choose 1 card from each of the 4 suits.
Ways = $^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} \times^{13} C_{1} = 13^{4} = 28561$ .

(iii) Four cards are face cards:
There are 12 face cards (J, Q, K of 4 suits) in a deck. We choose 4 from these 12.
Ways = $^{12} C_{4} = 495$ .

(iv) Two are red cards and two are black cards:
There are 26 red and 26 black cards. Choose 2 from each.
Ways = $^{26} C_{2} \times^{26} C_{2} = 325 \times 325 = 105625$ .

(v) Cards are of the same colour:
Either all 4 are red OR all 4 are black.
Ways = $^{26} C_{4} +^{26} C_{4} = 2 \times 14950 = 29900$ .

Total = $270, 725$ .
(i) $2860$ , (ii) $28561$ , (iii) $495$ , (iv) $105625$ , (v) $29900$ .

Q39

Find the numbers of words with or without meaning which can be made using all the letters of the word AGAIN. If all these words are arranged as in dictionary what will be the 49th word, 50th word?

Solution:

Step 1: Total words possible.
The word AGAIN has 5 letters: A, A, G, I, N. (A is repeated twice).
Total words = $\frac{5!}{2!} = 60$ .

Step 2: Dictionary arrangement.
Alphabetical order of letters: A, G, I, N.

Words starting with A: The remaining letters are A, G, I, N (all distinct).
Number of words = $4! = 24$ . (Words #1 to 24)
Words starting with G: The remaining letters are A, A, I, N.
Number of words = $\frac{4!}{2!} = 12$ . (Words #25 to 36)
Words starting with I: The remaining letters are A, A, G, N.
Number of words = $\frac{4!}{2!} = 12$ . (Words #37 to 48)

So, the first 48 words start with A, G, and I. The 49th word must be the first word starting with 'N'.

Step 3: Finding the 49th and 50th words.
Words starting with N use remaining letters: A, A, G, I.
The very first word (49th) alphabetically is when the remaining letters are sorted: A, A, G, I.
49th word = NAAGI
The next word (50th) is found by swapping the last two letters: G, I becomes I, G.
50th word = NAAIG

Total words = $60$ .
49th word = NAAGI.
50th word = NAAIG.

Q40

Determine the number of paths in the xy–plane from (1, 2) to (7, 5), where each such path is made up of individual steps going one unit to the right (R) or one unit upwards (U).

Solution:

Step 1: Calculate total steps needed.
To go from $(1, 2)$ to $(7, 5)$ , the path must move:
Right (R) steps = $7 - 1 = 6$ steps.
Up (U) steps = $5 - 2 = 3$ steps.
Total steps = $6 + 3 = 9$ steps.

Step 2: Find the combinations.
Any valid path is a sequence of 9 steps, containing exactly 6 R's and 3 U's. The number of such sequences is the number of ways to choose 3 positions for the 'U's out of the 9 total positions (or 6 for 'R's).
Number of paths = $^{9} C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$ .

There are $84$ possible paths.

Q41

How many positive integers greater than 5,000,000 can be formed using the digits 2, 3, 3, 5, 5, 6, 8?

Solution:

Step 1: Identify valid starting digits.
The number 5,000,000 is a 7-digit number. We are provided exactly 7 digits: {2, 3, 3, 5, 5, 6, 8}.
For a 7-digit number to be greater than 5,000,000, its first digit must be 5, 6, or 8.

Step 2: Case-by-case analysis.

Case 1: The number starts with 5.
The remaining 6 digits are: {2, 3, 3, 5, 6, 8} (Notice there are two 3's).
Arrangements = $\frac{6!}{2!} = \frac{720}{2} = 360$ .
Case 2: The number starts with 6.
The remaining 6 digits are: {2, 3, 3, 5, 5, 8} (Notice two 3's and two 5's).
Arrangements = $\frac{6!}{2! 2!} = \frac{720}{4} = 180$ .
Case 3: The number starts with 8.
The remaining 6 digits are: {2, 3, 3, 5, 5, 6} (Notice two 3's and two 5's).
Arrangements = $\frac{6!}{2! 2!} = \frac{720}{4} = 180$ .

Step 3: Total sum.
Total integers = $360 + 180 + 180 = 720$ .

There are $720$ positive integers greater than 5,000,000.

Q42

The board of directors of a pharmaceutical company has 10 members. An upcoming stockholder's meeting is scheduled to approve a new president, vice president, secretary and a treasurer. How many different ways the four can be appointed.

Solution:

Step 1: Understand the positions.
We are assigning 4 distinct roles (President, VP, Secretary, Treasurer) to 4 people out of 10. Since the roles are distinct, the order of selection matters. This is a permutation problem.

Step 2: Apply the permutation formula.
We need to arrange 4 out of 10 people.
Number of ways = $^{10} P_{4}$

Step 3: Calculate the value.
$^{10} P_{4} = 10 \times 9 \times 8 \times 7 = 5040$

The four members can be appointed in $5040$ different ways.

Q43

CASE STUDY 1.— Investment Decisions
An investment banker finalises the list of specific entities worthy of investment in each of the following three financial instruments:
• 3 Private Limited Companies for direct equity investment
• 5 Mutual Fund Schemes
• 2 Banks for making Fixed Deposits
Questions:
(a) The banker decides to invest the entire fund in only one entity. In how many ways can this investment be made?
(b) The banker chooses to invest in one entity of each of the three instruments (one company, one mutual fund, and one bank). In how many different ways can this be done?
(c) The banker decides to split the fund equally between two different entities, but the two chosen entities must be from different types of instruments (for example: one company & one mutual fund, or one mutual fund & one bank, etc.). In how many ways can such a pair of entities be chosen?

Solution:

Let $C = 3$ (Companies), $M = 5$ (Mutual Funds), and $B = 2$ (Banks).

(a) Entire fund in only one entity:
By the Fundamental Principle of Addition, the total number of options available is the sum of all entities across all categories.
$Ways = 3 + 5 + 2 = 10$ .

(b) One entity of each instrument:
By the Fundamental Principle of Multiplication, the banker must select 1 from $C$ AND 1 from $M$ AND 1 from $B$ .
$Ways = 3 \times 5 \times 2 = 30$ .

(c) Two entities from different instruments:
The pair could be (Company & Mutual Fund) OR (Mutual Fund & Bank) OR (Company & Bank).
Ways to select Company and Mutual Fund = $3 \times 5 = 15$ .
Ways to select Mutual Fund and Bank = $5 \times 2 = 10$ .
Ways to select Company and Bank = $3 \times 2 = 6$ .
Total combinations = $15 + 10 + 6 = 31$ .

(a) $10$ ways
(b) $30$ ways
(c) $31$ ways

Q44

CASE STUDY 2 – World Cup Playing XI with Restrictions
A cricket squad has: 8 batsmen, 6 bowlers, 4 all-rounders.
Management must select a Playing XI satisfying the following conditions:
• Exactly 5 batsmen
• At least 3 bowlers
• At least 2 all-rounders
Questions:
(a) In how many ways can a valid playing XI be selected?
(b) In how many ways can a Playing XI be selected if exactly 5 batsmen are chosen, at most 4 bowlers are selected, and the remaining players are all-rounders?
(c) If the 11 selected players line up for the anthem but the captain must stand at one end and the vice-captain at the other, how many arrangements are possible?

Solution:

(a) Selecting a valid Playing XI:
We must select 11 players. Since exactly 5 Batsmen (B) must be selected, we need $11 - 5 = 6$ players from Bowlers (W) and All-Rounders (AR).
Conditions for the 6 remaining players: At least 3 W and at least 2 AR.
Valid combinations of (W, AR):

3 W and 3 AR: $^{6} C_{3} \times^{4} C_{3} = 20 \times 4 = 80$
4 W and 2 AR: $^{6} C_{4} \times^{4} C_{2} = 15 \times 6 = 90$

Total ways to select Bowlers & All-rounders =

80 + 90 = 170

.
Ways to select 5 Batsmen =

^{8} C_{5} = 56

.
Total valid XI =

56 \times 170 = 9520

(b) Another selection scenario:
Condition: Exactly 5 Batsmen, at most 4 Bowlers, remaining AR. Again we need 6 players from W and AR. Since max AR available is 4, we evaluate possible combinations for 6 spots:

2 W and 4 AR (Sum=6): $^{6} C_{2} \times^{4} C_{4} = 15 \times 1 = 15$
3 W and 3 AR (Sum=6): $^{6} C_{3} \times^{4} C_{3} = 20 \times 4 = 80$
4 W and 2 AR (Sum=6): $^{6} C_{4} \times^{4} C_{2} = 15 \times 6 = 90$

Total ways to select W and AR =

15 + 80 + 90 = 185

.
Total valid XI =

^{8} C_{5} \times 185 = 56 \times 185 = 10360

(c) Arranging 11 selected players:
The 11 players are lined up. The Captain and Vice-Captain must be at the two extreme ends.
The Captain and Vice-Captain can occupy the 2 ends in $2! = 2$ ways.
The remaining 9 players can be arranged in the middle 9 positions in $9!$ ways.
Total arrangements = $2 \times 9! = 2 \times 362880 = 725760$ .

(a) $9520$ ways
(b) $10360$ ways
(c) $725760$ arrangements

Class 11- Applied Mathematics-NCERT Handbook Solutions-Chapter-8 (Combinatorics)

Chapter 8: Combinatorics

Check Your Progress 8.1

Check Your Progress 8.2

Check Your Progress 8.3

Check Your Progress 8.4

Check Your Progress 8.5

Chapter 8: Combinatorics

Practice Exercise: Multiple Choice & Subjective Questions

Chapter 8: Combinatorics

Subjective Questions & Case Studies