NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.4

NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.4 Permutations and Combinations help students apply the concepts of permutations and combinations to solve advanced and mixed-type problems. This exercise focuses on selecting and arranging objects in different scenarios, strengthening both conceptual clarity and application skills.

Prepared according to the latest CBSE Class 11 Maths syllabus, Exercise 6.4 includes problems where students need to decide whether to use permutation or combination based on the situation. Regular practice of this exercise improves logical thinking, accuracy, and problem-solving speed for exams.

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations Exercise 6.4

Permutations and Combinations

Easy

Q.

If ⁿC₈ = ⁿC₂, find ⁿC₂.

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Permutations and Combinations

Medium

Q.

Determine n if
(i) ²ⁿC₃ : ⁿC₃ = 12 : 1
(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1

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Permutations and Combinations

Easy

Q.

How many chords can be drawn through 21 points on a circle?

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Permutations and Combinations

Medium

Q.

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

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Permutations and Combinations

Medium

Q.

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

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Permutations and Combinations

Medium

Q.

Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

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Permutations and Combinations

Medium

Q.

In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

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Permutations and Combinations

Easy

Q.

A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

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Permutations and Combinations

Medium

Q.

In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

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The solutions are explained in a clear and step-by-step format so students can easily understand the approach and apply the correct formulas in different types of questions.

1. If $^nC_8 = ^nC_2$

nC8​=nC2​, find $^nC_2$

nC2​.

Ans: Since, we know that

$^nC_a = ^nC_b$

nCa​=nCb​

$\Rightarrow a = b$

⇒a=b or

$n = a + b$

n=a+b

That is,

$$^{10}C_2 = \frac{10!}{2!(10-2)!}$$

10C2​=2!(10−2)!10!​

$$\Rightarrow \frac{10!}{2!8!} = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}$$

⇒2!8!10!​=2×1×8!10×9×8!​

Hence,

$^nC_2 = 45$

nC2​=45.

2. Determine $n$

$n$ if

(i) $^{2n}C_3 : ^nC_3 = 12 : 1$

2nC3​:nC3​=12:1

Ans:

$$\frac{^{2n}C_3}{^nC_3} = \frac{12}{1}$$

nC3​2nC3​​=112​

$$\Rightarrow \frac{(2n)!}{3!(2n-3)!} \times \frac{3!(n-3)!}{n!} = \frac{12}{1}$$

⇒3!(2n−3)!(2n)!​×n!3!(n−3)!​=112​

$$\Rightarrow \frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 12$$

⇒(2n−3)!(2n)(2n−1)(2n−2)(2n−3)!​×n(n−1)(n−2)(n−3)!(n−3)!​=12

Therefore,

$$\Rightarrow \frac{2(2n-1)(2n-2)}{(n-1)(n-2)} = 12$$

⇒(n−1)(n−2)2(2n−1)(2n−2)​=12

$$\Rightarrow \frac{4(2n-1)(n-1)}{(n-1)(n-2)} = 12$$

⇒(n−1)(n−2)4(2n−1)(n−1)​=12

$$\Rightarrow 2n - 1 = 3(n-2)$$

⇒2n−1=3(n−2)

$$\Rightarrow 3n - 2n = -1 + 6$$

⇒3n−2n=−1+6

Hence,

$n = 5$

n=5.

(ii) $^{2n}C_3 : ^nC_3 = 11 : 1$

2nC3​:nC3​=11:1

Ans:

$$\Rightarrow \frac{(2n)!}{3!(2n-3)!} \times \frac{3!(n-3)!}{n!} = \frac{11}{1}$$

⇒3!(2n−3)!(2n)!​×n!3!(n−3)!​=111​

$$\Rightarrow \frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 11$$

⇒(2n−3)!(2n)(2n−1)(2n−2)(2n−3)!​×n(n−1)(n−2)(n−3)!(n−3)!​=11

$$\Rightarrow \frac{2(2n-1)(2n-2)}{(n-1)(n-2)} = 11$$

⇒(n−1)(n−2)2(2n−1)(2n−2)​=11

Therefore,

$$\Rightarrow \frac{(2n-1)}{(n-2)} = \frac{11}{4}$$

⇒(n−2)(2n−1)​=411​

$$\Rightarrow 8n - 4 = 11n - 22$$

⇒8n−4=11n−22

$$\Rightarrow 3n = 18$$

⇒3n=18

Hence,

$n = 6$

n=6.

3. How many chords can be drawn through 21 points on a circle?

Ans: For drawing one chord a circle, only 2 points are required.
To know the number of chords that can be drawn through the given 21 points on a circle, the number of combinations have to be counted.

Therefore, the chords can be drawn through 21 points taken 2 as equal to each chord.

Thus, required number of chords

$$^{21}C_2 = \frac{21!}{2!(21-2)!}$$

21C2​=2!(21−2)!21!​

$$\Rightarrow \frac{21!}{2!19!} = \frac{21 \times 20}{2}$$

⇒2!19!21!​=221×20​

$$\Rightarrow 210$$

⇒210

4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Ans: The team which is selected is of 3 boys and 3 girls from 5 boys and 4 girls.
The team of 3 boys can be selected from 5 boys in

$^5C_3$

5C3​ different ways.
The team of 3 girls can be selected from 4 girls in

$^4C_3$

4C3​ different ways.

Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected

$$^5C_3 \times ^4C_3$$

5C3​×4C3​

$$\Rightarrow \frac{5!}{3!2!} \times \frac{4!}{3!1!}$$

⇒3!2!5!​×3!1!4!​

$$\Rightarrow \frac{5 \times 4 \times 3!}{3! \times 2} \times \frac{4 \times 3}{3!}$$

⇒3!×25×4×3!​×3!4×3​

$$\Rightarrow 10 \times 4$$

⇒10×4

$$\Rightarrow 40$$

⇒40

5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls each colour.

Ans: The number of balls from which we have to select 6 red balls, 5 white balls, and 4 blue balls is 9 balls. These balls have to be selected in such a manner that each selection of balls consists of 9 balls from each of the colour.

Here, given we have –

3 balls which can be opted from the 6 red balls in

$^6C_3$

6C3​ different ways.
3 balls which can be opted from the 5 white balls in

$^5C_3$

5C3​ different ways.
3 balls which can be opted from the 5 blue balls in

$^5C_3$

5C3​ different ways.

Thus, after applying the multiplication principle, required number of ways of selecting the 9 balls will be –

$$^6C_3 \times ^5C_3 \times ^5C_3$$

6C3​×5C3​×5C3​

$$\Rightarrow \frac{6!}{3!3!} \times \frac{5!}{3!2!} \times \frac{5!}{3!2!}$$

⇒3!3!6!​×3!2!5!​×3!2!5!​

$$\Rightarrow \frac{6 \times 5 \times 4 \times 3!}{3!3 \times 2} \times \frac{5 \times 4 \times 3!}{3!2 \times 1} \times \frac{5 \times 4 \times 3!}{3!2 \times 1}$$

⇒3!3×26×5×4×3!​×3!2×15×4×3!​×3!2×15×4×3!​

$$\Rightarrow 20 \times 10 \times 10 = 2000$$

⇒20×10×10=2000

6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Ans: We have a deck of 52 cards, which contains 4 aces. When we make a combination, 5 cards should be made in such a manner that there is exactly one ace.

Now, 1 ace can be opted in

$^4C_1$

4C1​ different ways and the cards which are left, out of them 4 cards can be opted out of the 48 cards in

$^{48}C_4$

48C4​ different ways.

Therefore, after applying multiplication principle, the required number of 5 card combinations will be –

$$^{48}C_4 \times ^4C_1$$

48C4​×4C1​

$$\Rightarrow \frac{48!}{4!44!} \times \frac{4!}{1!3!}$$

⇒4!44!48!​×1!3!4!​

$$\Rightarrow \frac{48 \times 47 \times 46 \times 45 \times 44!}{44! \times 4 \times 3 \times 2 \times 1} \times \frac{4}{1}$$

⇒44!×4×3×2×148×47×46×45×44!​×14​

$$\Rightarrow \frac{48 \times 47 \times 46 \times 45}{3 \times 2}$$

⇒3×248×47×46×45​

$$\Rightarrow 778320$$

⇒778320

7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Ans: The number of players out of which we have to select is 17 players and only 5 players from them are bowlers.
A cricket team of 11 players need to be opted in such a way that there remains exactly 4 players which are bowlers.

4 bowlers can be opted in

$^5C_4$

5C4​ different ways whereas out of the remaining 7 players can be opted out from the 12 players in

$^{12}C_7$

12C7​ different ways.

Therefore, after applying multiplication principle, the required number of ways for selecting the cricket team will be –

$$^5C_4 \times ^{12}C_7$$

5C4​×12C7​

$$= \frac{5!}{4!1!} \times \frac{12!}{7!5!}$$

=4!1!5!​×7!5!12!​

$$\Rightarrow 5 \times \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}$$

⇒5×5×4×3×2×112×11×10×9×8​

$$= 3960$$

=3960

8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Ans: There are 5 black and 6 red balls in the bag. 2 black balls can be selected out of 5 black balls in

$^5C_2$

5C2​ ways and 3 red balls can be selected out of 6 red balls in

$^6C_3$

6C3​ ways.

Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls

$$^5C_2 \times ^6C_3$$

5C2​×6C3​

$$= \frac{5!}{2!3!} \times \frac{6!}{3!3!}$$

=2!3!5!​×3!3!6!​

$$\Rightarrow \frac{5 \times 4}{2} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}$$

⇒25×4​×3×2×16×5×4​

$$= 10 \times 20$$

=10×20

$$\Rightarrow 200$$

⇒200

9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Ans: There are 9 courses available out of which, 2 specific courses are compulsory for every student.

Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be chosen in

$^7C_3$

7C3​ ways.

Thus, required number of ways of choosing the programme is

$$^7C_3 = \frac{7!}{3!4!}$$

7C3​=3!4!7!​

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

⇒3×2×1×4!7×6×5×4!​=35

FAQs – Class 11 Maths Chapter 6 Exercise 6.4 Permutations and Combinations

Q1. What is the focus of Exercise 6.4?
Exercise 6.4 focuses on applying permutations and combinations in mixed problems where students must identify the correct method.

Q2. How do you decide whether to use permutation or combination?

• Use Permutation when order matters

• Use Combination when order does not matter

Q3. Why is Exercise 6.4 important for exams?
This exercise includes application-based questions that are commonly asked in exams and require strong conceptual understanding.

Q4. How can students prepare effectively for Exercise 6.4?
Students should practice different types of problems, understand the difference between permutation and combination clearly, and revise all formulas thoroughly.