^{n}C_{2} = ^{n}C_{7}.
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.
^{n}C_{2} = ^{n}C_{7}.
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.
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.