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Let $n, k$ be two positive integers such that $n>k$. Why is ${n\choose k}$ always divisible by a prime dividing $n$ (or even a product of such primes)? Please help me understand why. I cannot seem to wrap my mind around it. For instance, $ 30 \choose k $ is divisible by either $2,3,5$ or a product of those numbers: $$2\cdot3=6$$ $$2\cdot5=10$$ $$3\cdot5=15$$ etc.

Daniel Fischer
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    So... Are you asking why $\forall n>k>0,,\text{gcd}\left{{n\choose k},\ n\right}\neq 1$ ? –  Jun 18 '15 at 11:30
  • Yes. It's the same idea. Why? –  Jun 18 '15 at 11:32
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    This question http://math.stackexchange.com/questions/702927/prove-that-gcdn-choose-i-n-choose-j1-1i-jn?rq=1 gives a proof of this fact (take $i = 1$). But I find it singularly unenlightening - it tells me that it's true, but not why it's true - and would love to see a different (probably more combinatorial) proof. – Christopher Jun 18 '15 at 11:47
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    Ramunjndscpl: Instead of knocking heads with others asking for clarifications it would be more productive to EDIT your question so that new readers don't need to guess what you mean. Like: click edit -> replace "prime of $n$" with the correct "prime factor of $n$". Yes, many of us are good at guessing what you mean, but you should still fix it so that people showing up later don't need to spend time figuring it out, too. – Jyrki Lahtonen Jun 18 '15 at 12:38

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If $p$ is a prime, then \begin{align*} \binom{pr}{ps} &\equiv \binom rs \pmod p \\ \text{and}\qquad \binom{pr}{s} &\equiv 0 \pmod p \qquad\text{if $p\nmid s$.} \end{align*} Applying this to your situation: since $n>k>0$, there is a prime $p$ that divides $n$ with higher exponent than it divides $k$; applying the first fact iteratively we can reduce to the case where $p$ doesn't divide $k$ at all, and in that case the second fact shows that $p$ divides the binomial coefficient too. For example, for $\binom{120}{60}$: $$ \binom{120}{60} = \binom{2\cdot2\cdot2\cdot15}{2\cdot2\cdot 15} \equiv \binom{2\cdot2\cdot15}{2\cdot15} \equiv \binom{2\cdot 15}{15} \equiv 0 \pmod 2 $$

As Jyrki pointed out in comments, the two facts above are consequences of Lucas' theorem; see this answer for a proof. (And here's a combinatorial proof (pdf, 3 pages) I wrote up a few years ago.)

Community
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Let $X$ denote the set of all size-$k$ subsets of $\{1, ..., n\}$, where $0 < k < n$. Then $C_n$, the cyclic group of order $n$, acts on $X$ via its action on $\{1, ..., n\}$. Consider the stabiliser $G$ of an element $A$ of $X$. This is a subgroup of $C_n$, so $|G|$ is a factor of $n$. But $G$ also acts on $A$, so $|G|$ is also a factor of $k$; thus, $|G|$ is a factor of $\gcd\{n, k\}$. So by orbit-stabiliser, $|Orb_{C_n}(A)| = |C_n|/|G|$ is divisible by $n/\gcd\{n, k\}$. But this is true for any $A$, and since $X$ is the disjoint union of the orbits of the action by $C_n$, $|X|$ is divisible by $n/\gcd\{n, k\}$.

Finally, since $k < n$, $\gcd\{n, k\} < n$, and so $n/\gcd\{n, k\}$ is a factor of $n$ greater than $1$ which divides $|X| = {n\choose k}$.

Christopher
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  • I apologise if this uses mathematics beyond what the OP knows. If so, treat this as a self-answer to the question I asked in the comments. – Christopher Jun 18 '15 at 12:43
  • that's way too advanced for me to understand. I thank you anyways for your intelligent input. –  Jun 18 '15 at 14:40