Number of positive integral solutions of $a+b+c+d+e=20$ such that $a<b<c<d<e$ and $(a,b,c,d,e)$ is distinct

If $a<b<c<d<e$ it is trivial that $a,b,c,d,e$ are distinct numbers, so you can shorten the title. We may set $a=x_1$ and $b=a+x_2$, $c=b+x_3$, $d=c+x_4$, $e=d+x_5$, so we are looking for the solutions of $$ 5x_1+4x_2+3x_3+2x_4+x_5 = 20 $$ with $x_1,\ldots,x_5\in\mathbb{N}^+$. By setting $x_k=1+y_k$, we are looking for the solutions of $$ 5y_1+4y_2+3y_3+2y_4+y_5 = 5 $$ with $y_1,\ldots,y_5\in\mathbb{N}$. The number of such solutions is the coefficient of $z^5$ in the Taylor series at the origin of $$ \frac{1}{(1-z^5)(1-z^4)(1-z^3)(1-z^2)(1-z)} $$ i.e. the number of integer partitions of $5$, $p(5)=\color{red}{7}$.