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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.

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

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  • $\begingroup$ Thanks a lot. It is so easy however I did not think in that way :( $\endgroup$ – ZFR Feb 19 '18 at 12:09

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