Binomial division Looks very easy, but I can't make it:
$s \geq 2$ and $w \geq 2$ are prime numbers. $k$ is a natural number and $k \leq \min \{s,w \}$
Show that $\binom{s+w}{k}-\binom{w}{k} - \binom{s}{k}$ can be divided by $sw$.
 A: As Zev mentions, we do need some bounds on $k$.  We have your statement for $0\leq k\leq \min(s,w)$.  Here I will prove it for $1\leq k\leq \min(s-1,w-1)$.  (The other cases are not much different)  The goal is to look at your expression modulo $s$ and modulo $w$ and show it is $0$ in both cases.
Proof:  Expanding the binomials we have  $$\frac{(s+w)(s+w-1)\cdots(s+w-k+1)}{k!}-\frac{s(s-1)\cdots (s-k+1)}{k!}-\frac{w(w-1)\cdots(w-k+1)}{k!}.$$ Then since $s,w$ are primes, $k!$ is invertible, so that modulo $s$ this is
$$\equiv \frac{(w)(w-1)\cdots(w-k+1)}{k!}-\frac{w(w-1)\cdots(w-k+1)}{k!}\equiv 0\pmod{s}.$$    
Similarly for $w$.  Hence the original expression is divisible by both $s$ and $w$.
Elaboration:  Because $1\leq k<s$, $k!$ will be relatively prime to $s$ so that dividing by $k!$ makes "sense." Then we can write $\frac{1}{k!}\equiv a \pmod{s}$ for some integer $a$.  Using this, the term $$\frac{s(s-1)\cdots (s-k+1)}{k!}\equiv as(s-1)\cdots (s-k+1),$$ and must be divisible by $s$.  The fraction $$\frac{(s+w)(s+w-1)\cdots(s+w-k+1)}{k!}\equiv a(s+w)(s+w-1)\cdots(s+w-k+1)$$ $$\equiv a(w)(w-1)\cdots(w-k+1)\pmod{s}$$ since each $s$ is equivalent to zero.
A: HINT $\ $ Scale by $\rm\:k!\:$ then use $\rm\ f(0) = 0\ \Rightarrow\ x\:y\ |\ f(x+y) - f(x) - f(y)\:$ for $\rm\:f(x)\in \mathbb Z[x]\:.$  
Your problem is the special case is $\rm\ f(x) = x\ (x-1)\:\cdots\:(x-k+1)\:,\ \ x = s,\ y = w\:.$
