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!


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

  • $\begingroup$ Thank you so much! I completely overlooked factorial expansion, stupid mistake. I really appreciate the help! How would you figure out all numbers that it satisfies from there? $\endgroup$ – Nick Corin May 9 '13 at 15:49
  • $\begingroup$ It is a linear equation if you expand it ($k^2$ terms cancel) and hence there is only one solution. $\endgroup$ – Milind May 9 '13 at 15:52
  • $\begingroup$ Ah right, thanks to both of you! $\endgroup$ – Nick Corin May 9 '13 at 15:53
  • $\begingroup$ @nickcorin, find the edited version $\endgroup$ – lab bhattacharjee May 9 '13 at 15:55
  • $\begingroup$ @milind: It is not true that there is only one solution, there are infinitely many solutions (to the equation in the question, not to the linear equation). $\endgroup$ – Marc van Leeuwen May 12 '13 at 19:40

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 $n=7$.


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\}. $


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