Prove binomial coefficient

Question

Equality of Binomial coefficients. I was wondering why these two Binomial coefficients (x is just a place holder):

$\binom{x}{ k!(n+1-k)!}$ = $\binom{x}{k!(n-k)!}$. Both lead to $\binom{x}{k}$. Does the answer come from Pascal's triangle? I don't get it. Why is it equal?

Answer 1

This is not an identity.

For example, if $x=6, k=1, n=3$ then I would have thought $\binom{x}{ k!(n+1-k)!}=\binom{6}{ 1!\times3!}=\binom{6}{ 6}=1$ while $\binom{x}{ k!(n-k)!}=\binom{6}{ 1!\times2!}=\binom{6}{ 2}=15$ and $\binom{x}{k}=\binom{6}{1}=6$

So let's instead treat $\binom{x}{ k!(n+1-k)!} = \binom{x}{ k!(n-k)!}$ as an equation.

This requires one of:

• $x < k!(n-k)!$ and $n \ge k$
  • which would give $0=0$. That is not particularly interesting
• $x=k!$ and $n=k$
  • so $k!(n+1-k)! = k!(n-k)!=k!$ to give $1 = 1$
• $x=k!(n-k)!(n+2-k)$ and $n \ge k$
  • so $k!(n+1-k)! = x - k!(n-k)!$ to give $\binom{k!(n-k)!(n+2-k)}{ k!(n+1-k)!} = \binom{k!(n-k)!(n+2-k)}{ k!(n-k)!}$