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

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!

= 12×11/2×1

= 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!

              =12×11/2

              =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)=57×7!/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

My Personal Notes arrow_drop_up

Save