Counting Combinatoric Coefficients

We first recall a very useful formula for the power of a prime in a factorial. For every prime $p$ the power of $p$ in $n!$ is given by the following formally infinite sum: $\sum \limits_{i=1}^{\infty} \left \lfloor \frac{n}{p^i} \right \rfloor,$ where $\lfloor x\rfloor$ denotes the floor function.

We will prove that the number $\frac{(2n)!(2k)!}{n!k!(n+k)!}$ is always an integer by showing that for every prime $p$ its power in the numerator is greater than or equal to its power in the denominator. This follows from the following observation:

Lemma. For any two real numbers, $x$ and $y$ , $\lfloor 2x \rfloor + \lfloor 2y \rfloor - \lfloor x \rfloor - \lfloor y \rfloor - \lfloor x+y \rfloor \geq 0$

Proof of the Lemma. Because without the floor brackets the above sum is zero, the above inequality is equivalent to the reverse inequality for the fractional parts: $\{2x\} +\{2y\} -\{x\} -\{y\} -\{x+y\} \leq 0$

The left hand side is double-periodic, so we can assume that both $0\leq x <1$ and $0\leq y <1$ . Several cases have to be considered, depending on whether $2x$ , $2y$ and $x+y$ are less than 1 or greater than or equal to 1. It turns out that in all cases the left hand side is either -1 or 0.

The result now follows from the above Lemma, applied to $x=\frac{n}{p^i}$ and $y=\frac{k}{p^i}$ for all $i$ .

If you put any positive integers n and k, (2n)!(2k)! / n!k!(n+k)! will be an integer number because n!k!(n+k)! | (2n)!(2k)!. So, you can choose every pair for n and k. In this question, you are asked for the positive integers less than or equal then 20. (Combination 20 -> 1)(Combination 20 -> 1) = (20)(20) = 400