Finding $\binom{n}{4}$ if $\binom{n}{6}=\binom{n}{9}$ I was trying to solve a question related to combinatorics as follows :

A group contains $n$ persons. If the number of ways of selecting $6$ persons is equal to the number of ways of selecting $9$ persons, then find the number of ways of selecting $4$ persons from the group.

I dont have a step by step solution for concluding that $n=15$.
I tried to solve for $n$ but I am stuck at $(n-9)(n-8)(n-7)=(1/504)$

If $n<6$ then the number of ways to select $6$ people is the same as the number of ways to select $9$ people; in both cases this is $0$. For $n<6$ we can easily compute $\binom{n}{4}$:
$$\binom{5}{4}=5,\qquad\binom{4}{4}=1,$$
and $\binom{n}{4}=0$ when $n<4$.
If $n>6$, then by selecting $6$ people we leave $n-6$ people out. So the number of ways to select $6$ people is the same as the number of ways to select $n-6$ people. This suggests that $n-6=9$ and so $n=15$, and $\binom{15}{4}=1365$.
