Class 11 RD Sharma Solutions- Chapter 17 Combinations- Exercise 17.1 | Set 1

Chapter 17 of RD Sharma’s Class 11 Mathematics textbook focuses on Combinations a crucial topic in combinatorics. Combinations involve selecting items from the larger set where the order of the selection does not matter. This chapter aims to develop an understanding of how to calculate and analyze different ways of choosing items that have practical applications in probability, statistics, and various problem-solving scenarios.

Combinations

The Combinations refer to the selection of the items from a set where the order does not matter. For a set of n items the number of ways to choose r items is given by the combination formula:

where n! denotes the factorial of n. This formula is fundamental for solving problems involving the selection and arrangement where the sequence of the choices is irrelevant.

Question 1. Evaluate the following:

i) ¹⁴C₃

Solution:

We know that    ⁿC_r=n!/(n-r)!r!

=>¹⁴C₃=14!/(14-3)!3!

            =14!/11!3!

            =14x13x12/3x2x1

            =364

ii) ¹²C₁₀ 

Solution:

= 12!/(12-10)!10!

= 12!/2!10!

= 12x11/2x1

= 66

iii) ³⁵C₃₅

Solution:

= 35!/(35-35)!35!

= 1

iv) ⁿ⁺¹C_n

Solution:

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

= (n+1)!/n!

= n+1

v) 5

Solution:

∑ ⁵C_r=⁵C₁+⁵C₂+⁵C₃+⁵C₄+⁵C₅

r = 1

= 5+10+10+5+1

= 31

Question 2. If ⁿC₁₂=ⁿC₅, find the value of n.

Solution:

Given that ⁿC₁₂=ⁿC₅.

We know that two combinations will be equal when the sum of their r's is equal to n.

=>n=12+5=17.

Question 3. If ⁿC₄=ⁿC₆ , find ¹²C_n.

Solution:

=>n=6+4=10

=>¹²C₁₀=12!/10!2!

              =12x11/2

              =66

Question 4. If ⁿC₁₀=ⁿC₁₂ , ²³C_n.

Solution:

n = 10+12=22

=>²³C₂₂ = 23!/22!1!

              = 23

Question 5. If ²⁴C_x=²⁴C_2x+3 , find x.

Solution:

24 = x+2x+3

24 = 3x+3

21 = 3x

x = 21/3

x = 7

Question 6. If ¹⁸C_x=¹⁸C_x+2 , find x.

Solution: 

18 = x+x+2

18 = 2x+2

16 = 2x

x = 8

Question 7. If ¹⁵C_3r=¹⁵C_r+3, find r.

Solution: 

15 = 3r+r+3

15 = 4r+3

12 = 4r

r = 3

Question 8. If ⁸C_r-⁷C₃=⁷C₂, find r.

Solution: 

Given ⁸C_r-⁷C₃=⁷C₂

=>⁸C_r=⁷C₂+⁷C₃

We know that ⁿC_r+ⁿC_r-1=ⁿ⁺¹C_r

=>⁸C_r=⁸C₃

=>r=3

Question 9. If ¹⁵C_r:¹⁵C_r-1 = 11:5, find r.

Solution: 

¹⁵C_r/¹⁵C_r-1=11/5

(15!/(15-r)!r!)/(15!/(15-r+1)!(r-1)!)=11/5

15-r+1/r = 11/5

5(16-r) = 11r

80-5r = 11r

16r = 80

r = 5

Question 10. If ⁿ⁺²C₈:^n-2P₄=57:16, find n.

Solution:

We know that ⁿP_r=n!/(n-r)!

=>((n+2)!/(n+2-8)!8!)/((n-2)!/(n-2-4)!)=57/16

=>(n+2)(n+1)(n)(n-1)/8!=57/16

=>(n-1)n(n+1)(n+2)=(57/16)8!

=>(n-1)n(n+1)(n+2)=57x7!/2

=>(n-1)n(n+1)(n+2)=57x7x6x5x4x3

=>(n-1)n(n+1)(n+2)=19x3x7x6x5x4x3

=>(n-1)n(n+1)(n+2)=19x18x20x21

=>n=19

Read More:

• Combinations
• Combinations Formula with Examples

Conclusion

Understanding combinations is essential for the solving problems in probability and statistics where the order of the selection is not a factor. Mastery of the combination formula and its applications allows for the accurate calculation of possible outcomes in various scenarios enhancing problem-solving skills in the mathematics.