Factorial Equality Problem I'm stuck on this problem, any help would be appreciated.
Find all $n \in \mathbb{Z}$ which satisfy the following equation:
$${12 \choose n} = \binom{12}{n-2}$$
I have tried to put each of them into the factorial equation and then making them equal each other and manipulating a bit, although I get stuck at:
$$(12 - k)!\cdot k! - (14 - k)!\cdot (k - 2)! = 0$$
Maybe I'm on the completely wrong path? Maybe I just had a mathematical error?
Thanks in advance!
 A: Which reduces to $$(12-k)! \cdot k(k-1) \cdot (k-2)!=(14-k)(13-k)\cdot (12-k)!\cdot (k-2)!$$
$$\implies k(k-1)=(14-k)(13-k)$$

Alternatively, we know $$\binom nr=\binom n{n-r}$$ and $\binom nr\ne \binom ns$ for $s\ne r,n-r$ 
So, $n+(n-2)=12$
A: A better way is to recognize that:
$$\binom{12}{6-j} = \binom{12}{6+j}$$
for $j=0,1,..6$, which is to say that the binomial coefficients are symmetric about the middle coefficient.
So here $n$ and $n-2$ have to be on either side of $6$, and equal distances away. This means their average, $n-1$ has to be the center and hence equal to $6$, giving $n=7$.
A: The nonzero binomial coefficients with upper index $12$ are given by line $12$ in Pascal's triangle:
$$
  1,12,66,220,495,792,924,792,495,220,66,12,1.
$$
Among these equality only occurs for entries that are symmetric around the central number$~\binom{12}6$, which gives the solution $n=7$ to your equation. However all other numbers $\binom{12}k$ are zero, which gives infinitely many more solutions to you equation, namely for all integers $n<-1$ and $n>14$.
${}$
All in all the set of solutions is:
$
  \Bbb Z \setminus\{0,1,2,3,4,5,6,8,9,10,11,12,13,14\}.
$
