Class 11 Maths Chapter 6 Permutations and Combinations – Exercise 6.2 NCERT Solutions

Introduction

Exercise 6.2 introduces the concept of combinations. Unlike permutations, combinations focus on selections where order does not matter. You will learn how to calculate the number of ways to choose objects from a set and apply these ideas to practical problems.

Key Concepts

1. Combination Formula: Number of combinations of $n$ objects taken $r$ at a time:

$$C(n,r)=\left(\genfrac{}{}{0px}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$$

2. Properties of Combinations:

   • $\left(\genfrac{}{}{0px}{}{n}{0}\right)=1$

   • $\left(\genfrac{}{}{0px}{}{n}{n}\right)=1$

   • $\left(\genfrac{}{}{0px}{}{n}{r}\right)=\left(\genfrac{}{}{0px}{}{n}{n-r}\right)$

3. Difference from Permutations:

   • Permutations: order matters.

   • Combinations: order does not matter.

NCERT Questions with Solutions (10)

Q1. Find number of combinations of 5 objects taken 3 at a time.

$$C(5,3)=\frac{5!}{3!\cdot 2!}=\frac{120}{12}=10$$

Q2. Find number of combinations of 6 objects taken 2 at a time.

$$C(6,2)=\frac{6!}{2!\cdot 4!}=\frac{720}{48}=15$$

Q3. Find number of combinations of 4 objects taken all at a time.

$$C(4,4)=1$$

Q4. Find number of combinations of 7 objects taken 1 at a time.

$$C(7,1)=7$$

Q5. Find number of combinations of 8 objects taken 3 at a time.

$$C(8,3)=\frac{8!}{3!\cdot 5!}=\frac{40320}{720}=56$$

Q6. Find number of combinations of 9 objects taken 2 at a time.

$$C(9,2)=\frac{9!}{2!\cdot 7!}=\frac{362880}{10080}=36$$

Q7. Find number of combinations of 10 objects taken 4 at a time.

$$C(10,4)=\frac{10!}{4!\cdot 6!}=\frac{3628800}{17280}=210$$

Q8. Find number of combinations of 12 objects taken 5 at a time.

$$C(12,5)=\frac{12!}{5!\cdot 7!}=\frac{479001600}{604800}=792$$

Q9. Find number of combinations of 11 objects taken 3 at a time.

$$C(11,3)=\frac{11!}{3!\cdot 8!}=\frac{39916800}{241920}=165$$

Q10. Find number of combinations of 6 objects taken 0 at a time.

$$C(6,0)=1$$

FAQs (10)

1. What is combination? Selection of objects without regard to order.

2. What is formula for combination? $C(n,r)=\frac{n!}{r!(n-r)!}$.

3. What is $\left(\genfrac{}{}{0px}{}{n}{0}\right)$? Always equal to 1.

4. What is $\left(\genfrac{}{}{0px}{}{n}{n}\right)$? Always equal to 1.

5. What is relation between $\left(\genfrac{}{}{0px}{}{n}{r}\right)$ and $\left(\genfrac{}{}{0px}{}{n}{n-r}\right)$? They are equal.

6. Is order important in combination? No.

7. Is repetition allowed in combination? Not in standard formula.

8. What is difference between permutation and combination? Permutation considers order, combination does not.

9. What is domain of factorial function? Non‑negative integers.

10. Why is this exercise important? It builds foundation for probability and binomial theorem.

Conclusion

Exercise 6.2 has 10 solved questions and 10 FAQs that strengthen your understanding of combinations. This completes the Permutations and Combinations chapter in Class 11 Maths.

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New Syllabus-Class 11 Maths Chapter 6 Permutations and Combinations – Exercise 6.2 NCERT Solutions