Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking for a combinatorial proof to the following statement:

$$ p\mid\binom{p}{k} \ \ \ , \ \ 0<k<p \ \ \ \ \ \ \text{and} \ \ p \ \text{is prime}.$$

Thank you.

share|cite|improve this question
up vote 5 down vote accepted

Consider the set of all $f \colon {\mathbb Z}_p \to \{0,1\}$ which take 1 at exactly $k$ points. This set has $p \choose k$ elements. You can think of putting 0-1's in a circle.

Any such labelling can be rotated in exactly $p$ different ways, so the cardinality of the set divides $p$.

[If rotating did not change a labelling, then $f(x+i)=f(x)$ for some $i$. Because $i,p$ are coprime, $ai+bp=1$ for some integers; therefore $f(x)=f(x+ai)=f(x+1-bp)=f(x+1)$ and the function is constant, but this is impossible when $0<k<p$.]

A similar proof shows that $a^p \equiv a \pmod p$.

share|cite|improve this answer
Very nice! Thank you! – Salech Alhasov Apr 3 '12 at 10:12

If you count the number of $k$-element subsets of a $p$-element set that contain a given element and then sum over all $p$ elements, you get $k\binom pk$, since each element is in $k$ of the $\binom pk$ subsets. This is $p$ times the count for a single element, and since $p\nmid k$ for $0\lt k\lt p$, the factor $p$ must be in $\binom pk$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.