Factorial Equality Problem

Question

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!

Answer 1

Which reduces to $$(12-k)! \cdot k(k-1) \cdot (k-2)!=(14-k)(13-k)\cdot (12-k)!\cdot (k-2)!$$

$$\implies k(k-1)=(14-k)(13-k)$$

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Alternatively, we know $$\binom nr=\binom n{n-r}$$ and $\binom nr\ne \binom ns$ for $s\ne r,n-r$

So, $n+(n-2)=12$

Answer 2

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 $n=7$.

Answer 3

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 $n>14$.

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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\}. $