Is $37!$ divisible by $19!\cdot20!$ ?
I rewrite it as $\frac{37!}{19!*20!}=\frac{37*36*35*...*21}{19!}$
simplified it and got $\frac{37*35*33*31*29*23}{19}$ what is obvious is not integer.
Another way, I noticed that fraction $\frac{37*36*35*...*21}{19!}$ numerator doesn't have a multiple of $19$, so $19\nmid 37*36*35*...*21$
But I believe that it should be more general way to prove it. Any ideas? Thanks.
$$ \frac {37!}{19!20!}$$ is not an integer
We know that there are two $19$ in the bottom but only one in the top.
Since $19$ is prime the fraction does not simplify to an integer.