# Thread: Find n such that (related to combinations)...

1. ## Find n such that (related to combinations)...

nC2 = nC7.

n! / 2!(n - 2)! = n! / 7!(n - 7)!

You can divide n! out from both sides.

However, I reached...

1 / 2! = (n - 2)(n - 3)(n - 4)(n - 5)(n - 6) / 7!

Something seems off. Too many terms are involved for my liking.

2. ## Re: Find n such that (related to combinations)...

Remember that the binomial coefficients, $^nC_i$, are symmetric- $^3C_i$= 1, 3, 3, 1 and $^4C_i$= 1, 4, 6, 4, 1, etc.

In order that $^nC_j= ^nC_k$, we must have j= n- k.