TITLE: Combinatorics question: Show divisibility
QUESTION [7 upvotes]: Let $a\geq2$, $b\geq2$ be two prime numbers and k be a natural number with $k\leq min(a,b)$.
How can one show that $z := \binom{a+b}{k} - \binom{a}{k} - \binom{b}{k}$ is divisible by the product $ab$?

REPLY [2 votes]: Let $A$ and $B$ be disjoint sets of size $a$ and $b$, and let $S$ be the sets of subsets of $A\cup B$ of size k which contain at least one element of $A$ and one element of $B$.
Then we can find an action of the additive group $\mathbb{Z}_a\oplus\mathbb{Z}_b$ on $S$ which acts by rotating the elements of $A$ and $B$ separately.  You can make a pretty easy argument that each of the orbits of this action on $S$ has size $ab$.  (This is where you need the upper limit on $k$. When $k\leq a$, the subset cannot include all of $A$, and likewise for $b$ and $B$.) so you can show that the size of $S$ must be a multiple of $ab$.  But the size of $S$ is just:
$${{a+b} \choose {k}} - {{a}\choose{k}}- {{b}\choose{k}}$$
when $\operatorname{min}(a,b)\geq k>0$.
Similar results:
If $a\leq k \leq b$ then:
$${{a+b} \choose {k}} - {b \choose {k-a}} - {b \choose k}$$
is divisible by $ab$.
If $a+b > k \geq \operatorname{max}(a,b)$ then:
$${{a+b} \choose {k}} - {b \choose {k-a}} - {a \choose {k-b}}$$
is divisible by $ab$.