Combinatorial interpretation of this number?

Question

It is straightforward to show that if $m,n\in\mathbb{Z}$ and $m\geq n$, then $$m\mid \gcd(m, n)\binom{m}{n}.$$ I'm trying to find a combinatorial interpretation of this fact, but I can't seem to come up with one. My two proofs are formal, and do not give me any combinatorial insight.

Well the binomial coefficient can also be represented as $\frac{m!}{n!(m-n)!}$, so then you can see that m divides that since $\gcd(m,n)$ is an integer. — kvmu, Jun 30 '13 at 20:42
@kvmu The OP is asking for a combinatorial insight. E.g.: Show that $\binom{n}{k}$ is an integer. Proof: Well, it is the number of $k$ element subset of an $n$ element set. That's gotta be an integer. — Lord Soth, Jun 30 '13 at 20:43

score 4 · Accepted Answer · answered Jun 30 '13 at 20:44

4

$\mathbb Z/m$ acts (by transitions) on the set of $n$-element subsets of $\mathbb Z/m$. The action is not free but its restriction on any subgroup of order coprime to $n$ — in particular, on $\gcd(m,n)\mathbb Z/m\subset\mathbb Z/m$ — is free. So $\frac m{\gcd(m,n)}\mid\binom mn$.

answered Jun 30 '13 at 20:44

Grigory M

18,231

(Oh, and the situation is especially clear when $m$ is prime.) – Grigory M Jun 30 '13 at 20:47

score 3 · Answer 2 · answered Jun 30 '13 at 20:49

Set $\gcd(m,n)=d$ and write $m=da, n=db$ where $\gcd(a,b)=1$. Then your statement is equivalent to $$a\left|{da\choose db}\right.$$

Consider subsets of size $db$ from $S=\{0,1,2,\ldots, da-1\}$. The permutation $\phi:S\rightarrow S$ given by $\phi(x)=x+d\pmod{da}$. $\phi$ also induces a map on subsets of size $db$. We may call two such subsets equivalent if we can get from one to another by iterating $\phi$. Iterating $\phi$ $a$ times we get back to where we started, and in fact all the $a$ subsets must be distinct because $\gcd(a,b)=1$. Hence the subsets of size $db$ have been partitioned into classes of size $a$, which proves the statement.

(Of course this is exactly the same proof — but maybe this description is easier to understand.) — Grigory M, Jun 30 '13 at 20:56
@GrigoryM Sorry, I didn't see yours until I'd already posted mine. — vadim123, Jun 30 '13 at 20:58

score 1 · Answer 3 · answered Jun 30 '13 at 21:07

This isn't entirely combinatorial, but:

We prove that $\begin{pmatrix} m \\ n \end{pmatrix}$ is always a multiple of $m/\gcd(m,n)$ using the left multiplication action of $\mathbb{Z}_m$ on the set $X$ of all its subsets of size $n$. By definition, $|X|=\begin{pmatrix}m\\n\end{pmatrix}$. By the theory of group actions, the set $X$ can be partitioned into orbits. Then it is enough to show that every orbit must have size that is a multiple of $m/\gcd(m,n)$. By the orbit stabilizer theorem, then, it is enough to prove that the stabilizer of a set $E$ of size $n$ must have a size that divides $\gcd(m,n)$.

Let $E$ be a subset of $\mathbb{Z}_m$ of size $n$. The stabilizer of $E$ is a subgroup of $\mathbb{Z}_n$, and all subgroups of $\mathbb{Z}_m$ are cyclic. Furthermore, if we choose the minimal member $d$ of any subgroup of $\mathbb{Z}_m$, then $d$ generates that subgroup. So let $d$ be the minimal generator of the stabilizer of $E$. So the stabilizer of $E$ has size $m/d$, which means that we need to show that $m/d$ divides $\gcd(m,n)$.

But if we choose an element $x$ of $E$, then the elements $x,x+d,x+2d,\dots$ all belong to $E$ as well (by the definition of a stabilizer); i.e., if $H$ is the stabilizer of $E$, then the coset $x+H$ is a subset of $E$. So $E$ is a union of cosets of $H$, all of which have size $m/d$. So $n$ is a multiple of $m/d$, which means that $m/d$ is a common factor of $m$ and $n$. So $\gcd(m,n)$ divides $m/d$, as we wanted.

Combinatorial interpretation of this number?

3 Answers3