<s>[INST] 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
$$\binom{9}{6}=\binom{3\cdot 3}{2\cdot 3}=\frac{9\cdot8\cdot7}{3\cdot 2\cdot 1}=4(3\cdot 7)$$
$$\binom{16}{8}=\binom{4\cdot 4}{2\cdot 4}=\frac{16\cdot15\cdot...\cdot 10\cdot 9}{8\cdot 7\cdot...\cdot2\cdot 1}=5(2\cdot 3^2\cdot11\cdot 13)$$
 [/INST] 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.
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