How can I solve for $n$ in this binomial coefficient equation

Question

How can I solve for $n$ in this binomial coefficient equation? $${n\choose 3} = {n\choose 9}$$

When I try to expand it using factorials, I get a very, very long equation, involving $n-s$ up to $n^6$

I also know that $n = 12$;

Answer 1

You professor or book is trying to get you to think about the relation that $$ \binom{n}{m} = \binom{n}{n-m}$$

The meaning is that choosing $m$ objects out of $n$ is the same as selecting the $n-m$ objects not to choose.

Thus in your case, $n=12$.

Answer 2

$$ \binom{n}{3} = \binom{n}{9} $$ requires $$ \frac{9!}{3!} = \frac{(n-3)!}{(n-9)!} $$ note that $n=12$ is a solution, and whereas LHS is constant, RHS is a strictly increasing function of $n$, so this is the only solution