The answer is the coefficient of the $x^{20}$ in the expansion of the following polynomial $$(x+x^2 +\cdots + x^{20})(x^2+x^4+\cdots+x^{20})(x^5+x^{10}+x^{15}+x^{20})$$

As the involved numbers are relatively small, this problem can also be solved using casework. Note that given any legitimate combination of $5$ and $2$, there always exists a unique count of $1$ such that their sum equals $20$.

It is clear that $5$ can only appear $1$, $2$, or $3$ times.

• When $5$ appears $1$ times, $2$ can appear $1$ to $7$ times, inclusive. There are totally $7$ legitimate combinations.
• When $5$ appears $2$ times, $2$ can appear $1$ to $4$ times, inclusive. There are totally $4$ legitimate combinations.
• When $5$ appears $3$ times, $2$ can appear either $1$ or $2$ times. There are totally $2$ legitimate combinations.

Hence, the final answer is $7+4+2=\boxed{13}$.