0

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.

RFZ
  • 16,814
  • What does your notation mean? Did you possibly mean $\binom {2p-q-1}{p-q}$? – lulu Feb 19 '18 at 12:02
  • 1
    The numerator must contain $p$ and the nominator can't. Just write it down then you will understand the answer from @tong_nor. – Raphael J.F. Berger Feb 19 '18 at 12:03
  • @lulu, Yes, I did. This is a notation of binomial coefficient. – RFZ 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 :) – RFZ Feb 19 '18 at 12:05
  • Never knew that. I suggest editing your post to make it clear...I expect most readers here will interpret it the way I did. – lulu Feb 19 '18 at 12:06
  • @lulu, Also $\tbinom{n}{k}=C_n^k$ – RFZ Feb 19 '18 at 12:06
  • As I say, that is not the standard notation in English speaking countries. As you can see from the posted solution below, your notation is going to confuse your readers. – lulu Feb 19 '18 at 12:07
  • @lulu, I've corrected it. – RFZ Feb 19 '18 at 12:08
  • Given the non-standard interpretation, the claim is trivial. $p<2p-q-1$ so $p$ divides the numerator. $p>p-q$ and $p>2p-q-1-(p-q)=p-1$ so $p$ does not divide the denominator. – lulu Feb 19 '18 at 12:09
  • @lulu, Yes indeed. It turns out to be easy fact :( – RFZ Feb 19 '18 at 12:10

1 Answers1

2

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 $2p-q-1\ge p$.

tong_nor
  • 3,994