Does $n+1$ divides $\binom{an}{bn}$? Suppose that $a>b>0$ be integers.  Is it true that for an integer $n>2$ that
$$n+1|\binom{an}{bn}$$
or is there a counter example.  Certainly i think the right hand side would reduce to
$$\frac{an(an-1)(an-2)...((a-1)n+1)}{n(n-1)(n-2)...2\cdot 1}$$
But I'm not seeing how this could reduce better to show there is a factor of $n+1$ left.  
Examples show this is true for small n; for example
$$\binom{9}{6}=\binom{3\cdot 3}{2\cdot 3}=\frac{9\cdot8\cdot7}{3\cdot 2\cdot 1}=4(3\cdot 7)$$
$$\binom{16}{8}=\binom{4\cdot 4}{2\cdot 4}=\frac{16\cdot15\cdot...\cdot 10\cdot 9}{8\cdot 7\cdot...\cdot2\cdot 1}=5(2\cdot 3^2\cdot11\cdot 13)$$
 A: There is a counterexample. 
Take $(n,a,b)=(3,5,1)$. 
$\binom{5\times 3}{1\times 3}=455$ is not divisible by $4$.
A: Here is a larger counterexample : $a=6,b=2,n=4$, then $n+1$ is not a multiple of $\binom{an}{bn}$.
Another one is given by $a=5,b=2,n=5$.
Also, note that $\binom{an}{bn} = \binom{an}{(a-b)n}$, therefore replacing $b$ above with $a-b$ would also work.
I am still thinking about $b > 2$(other than taking $b \to a-b$ obviously) though. I have not been able to find an example yet.
A: As counterexamples for every $n=p-1$ for any prime $p$ and considering $a=n+2=p+1$ and any $b$ with $0 \lt b \lt a$
Then $(an)! = (p^2-1)!$ is divisible by $p^{p-1}=p^n$ but not $p^{p}=p^{n+1}$, while $(bn)!(an-bn)!$ is divisible by $p^{b-1}p^{a-b-1}=p^{n}$, so ${an \choose bn}$ is not divisible by $p=n+1$
This gives for example the counterexamples 
n   a   b   C(an,bn)  n+1   
1   3   1          3    2
2   4   1         28    3
2   4   2         70    3
2   4   3         28    3
4   6   1      10626    5
4   6   2     735471    5
4   6   3    2704156    5
4   6   4     735471    5
4   6   5      10626    5
6   8   1   12271512    7
...     

