Question

If nC12=nC13, then find the value of 25Cn. (i)2 (ii)100 (iii)1 (iv)3

Answer 1

Answer:

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

1/10!(n-10)×(n-11)!×(n-12)!=1/12×11×10!×(n-12)!

1/(n-10)×(n-11)= 1/12×11

12= n^2--11n-10n+110

n^2-21n+98=0

after factorization....

(n-14). (n-7). =0

n=14,7

after put the value of n in....

23!/n!(23-n)!....1. put n=14...

.................. put n=7after solving you get value of 23cn....make me brainleast and follow me

Answer 2

Answer:

The value of 25Cn is 1.

Step-by-step explanation:

Given,

nC12 = nC13

n!/(n-12)! * 12! = n!/(n-13)! * 13!

1/(n-12)(n-13)! * 12! = 1/(n-13)! * 13*12!

1/(n-12) = 1/13

n-12 = 13

n=25

Thus, nC25 = 25C25

                    = 25!/(25-25)! * 25!

                    = 1