# HOW TO FIND THE VALUE OF N AND R IN COMBINATION

How to Find the Value of n and r in Combination :

In this section, we will learn, how to find the values of n and r in the given combination.

## How to Find the Value of n and r in Combination - Questions

Question 1 :

If nC12 = nC9 find 21Cn.

Solution :

nC12  =  n!/(n - 12)! 12!  -----(1)

nC9  =  n!/(n - 9)! 9!     -----(2)

(1)  =  (2)

n!/(n - 12)! 12!  =  n!/(n - 9)! 9!

(n - 9)! 9!  =  (n - 12)! 12!

(n - 9)(n - 10)(n - 11)(n - 12)! 9!  =  (n - 12)! 12 ⋅ 11 ⋅ 10 ⋅ 9!

(n - 9)(n - 10)(n - 11)  =  12 ⋅ 11 ⋅ 10

n - 9  =  12

n  =  12 + 9

n  =  21

21Cn =   21C21  =  1

Question 2 :

If 15C2r−1 = 15C2r+4, find r.

Solution :

If nC=  nCy  ==>  x  =  y (or) x + y  =  n

2r - 1 + 2r + 4  =  15

4r + 3  =  15

4r  =  12

Divide by 4 on both sides.

r  =  12/4  ==>  3

Hence the value of r is 3.

Question 3 :

If nPr = 720, and nCr = 120, find n, r.

Solution :

nPr = 720

n!/(n - r)!  =  720  ---(1)

nCr = 120

n!/(n - r)! r!  =  120  ---(2)

Divide (1) by (2), we get

r!  =  720/120

r!  =  6

r!  =  3!  ==> r = 3

By applying the value of r in the (1), we get

n!/(n - 3)!  =  720

n(n - 1) (n - 2)  =  720

n(n - 1) (n - 2)  =  10 ⋅ 9 ⋅ 8

n  =  10

Hence the value of r and n are 3 and 10 respectively.

Question 4 :

Prove that 15C3 + 2 × 15C4 + 15C5 = 17C5.

Solution :

L.H.S

=   15C3 + 2 × 15C4 + 15C5

=   15C3 + 15C4 + 15C15C5

=   15C4 15C3 +  15C15C4

By using the property :

nCr + n Cr−1 = n+1Cr

=  16C4 16C5

=  17C5 ---> R.H.S

Question 5 :

Prove that 35C5 + ∑ 4r=0 (39−r)C4  =  40C5.

Solution :

L.H.S

=  35C5 + ∑ 4r=0 (39−r)C4

=  35C5 + 39C438C37C36C35C4

By using the property :

nCr + n Cr−1 = n+1Cr

=  35C5 +  35C4 + 39C438C37C36C

=  36C36C4 39C438C37C

=  37C37C39C438C

=  38C+ 38C 39C

=  39C 39C

=  40C

After having gone through the stuff given above, we hope that the students would have understood, how to find the values of n and r in the given combinations

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