For any prime number $p$ and natural number $i < p$, prove that $p$ divides ${p \choose i}$.
Also, what happens when $p$ is not a prime. Is this still true?
I tried writing out the formula for combination but couldn't get further.
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Martin Sleziak
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This question has been already asked before. I recommend you to search for the answers. On the other hand, if $ p $ isn't prime the statement is false. Just take for example $ p=6$ and $ i=3$. – Xam Nov 28 '16 at 02:33
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More generally $,\displaystyle\dfrac{\gcd(n,m)}n{n\choose m}, $ is an integer. OP is special case $,n,$ prime. – Bill Dubuque Nov 28 '16 at 03:23
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You should specify $0<i$, as “natural number” is far from unambiguously excluding $0$. – PJTraill Nov 30 '16 at 10:37
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By a combinatorial argument, or by manipulating factorials, we have $$i\binom{p}{i}=p\binom{p-1}{i-1}\ .$$ Since $\binom{p-1}{i-1}$ is an integer, $$p\mid i\binom{p}{i}\ .$$ But $p$ is prime and $1\le i<p$, so $p$ and $i$ have no common factor, so $$p\mid \binom{p}{i}\ .$$ For the case when $p$ is not prime, just take $p=4$ and try out various values of $i$.
David
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This method generalizes to $,\displaystyle\dfrac{\gcd(n,m)}n{n\choose m}, $ is an integer. OP is special case $,n,$ prime. – Bill Dubuque Nov 28 '16 at 03:25
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It is not true in general (look for a counterexample at $p=4$). For $p$ prime though, you have $$ {p\choose i} = \frac{p!}{i!(p-i)!} \in \mathbb{N} $$ So $$ p \mid {p\choose i}i!(p-i)! $$ However, $p\nmid i!, p\nmid (p-i)!$. So $p\mid {p\choose i}$.
Prahlad Vaidyanathan
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1You need to justify those indivisibilities - which are the crux of the matter. – Bill Dubuque Nov 28 '16 at 03:17
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If $k<p$, then $p\nmid j$ for each $1\leq j\leq k$. Since $p$ is prime, this implies $p\nmid k!$ as well (See https://en.wikipedia.org/wiki/Euclid's_lemma) – Prahlad Vaidyanathan Nov 29 '16 at 10:45