# Divisibility of binomial coefficient

Let $p$ be a prime number and $q\in \{1,\dots,p-1\}$. Prove that $\tbinom{2p-q-1}{p-q} \equiv 0\pmod {p}$

However, I have no idea how to prove this.

Would be thankful for solution.

• What does your notation mean? Did you possibly mean $\binom {2p-q-1}{p-q}$? – lulu Feb 19 '18 at 12:02
• The numerator must contain $p$ and the nominator can't. Just write it down then you will understand the answer from @tong_nor. – Rudi_Birnbaum Feb 19 '18 at 12:03
• @lulu, Yes, I did. This is a notation of binomial coefficient. – ZFR Feb 19 '18 at 12:03
• Well, it's not a standard notation. Normally $C^n_r=\binom nr$. – lulu Feb 19 '18 at 12:04
• @lulu, In Russian universities and books it is a standard notation :) – ZFR Feb 19 '18 at 12:05

Assuming $C^n_k=\binom{n}{k}$:
$2p-q-1>p-q$, so you just have $C^{p-q}_{2p-q-1}=0$ $\dots$
Also $C_{p-q}^{2p-q-1}=\frac{(2p-q-1)!}{(p-q)!(p-1)!}$ is divisible by $p$, since $p-q,\ p-1\in\{0,\dots,p-1\}$ and $2p-q-1\ge p$.