For any prime number $p$ and natural number $i < p$, prove that $p$ divides ${p \choose i}$.

Question

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.

Answer 1

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 $p\mid {p\choose i}$.

Answer 2

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 $i$.