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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!

=>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

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.

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