Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

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

  • Last Updated : 11 Feb, 2021

Question 1. Evaluate the following:

i) 14C3

Solution:

We know that    nCr=n!/(n-r)!r!

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

=>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+1Cn

Solution:

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

= (n+1)!/n!

= n+1

v) 5

Solution:

5Cr=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 12Cn.

Solution:

=>n=6+4=10

=>12C10=12!/10!2!

              =12×11/2

              =66

Question 4. If nC10=nC12 , 23Cn.

Solution:

n = 10+12=22

=>23C22 = 23!/22!1!

              = 23

Question 5. If 24Cx=24C2x+3 , find x.

Solution:

24 = x+2x+3

24 = 3x+3

21 = 3x

x = 21/3

x = 7

Question 6. If 18Cx=18Cx+2 , find x.

Solution: 

18 = x+x+2



18 = 2x+2

16 = 2x

x = 8

Question 7. If 15C3r=15Cr+3, find r.

Solution: 

15 = 3r+r+3

15 = 4r+3

12 = 4r

r = 3

Question 8. If 8Cr7C3=7C2, find r.

Solution: 

Given 8Cr7C3=7C2

=>8Cr=7C2+7C3

We know that nCr+nCr-1=n+1Cr

=>8Cr=8C3

=>r=3

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

Solution: 

15Cr/15Cr-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 nPr=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
Recommended Articles
Page :