Let $S_{a, b, c, d, e} $ be the number of solutions with $x_1\ge a$, $x_2 \ge b$, $x_3\ge c$, $x_4\ge d$ and $x_5\ge e$.
The equation can be written as
$$(x_1-a)+(x_2-b)+(x_3-c)+(x_4-d)+(x_5-e)=21-a-b-c-d-e$$
So we have
$$S_{a, b, c, d, e} =\binom{21-a-b-c-d-e+4}{4}$$
if $a+b+c+d +e\le21$ and is $0$ otherwise.
The answer to this question is
$$S_{0, 1,15,0,0}-S_{4,1,15,0,0}-S_{0,4,15,0,0}+S_{4,4,15,0,0}=\binom{9}{4}-\binom{5}{4}-\binom{6}{4}+0=106$$
As mentioned in the comments, there are three cases: $x_1\in \{1,3,5\}$. That's because the minimum value of $x_2+x_3+x_4+x_5$, when all are positive integers, is $4$ and so the maximum odd value of $x_1$ is $5$. If we set $x_1 =1$, then there are $9$ stars left and $8$ spaces to place $3$ bars since all four remaining variables must be positive integers. This gives ${8\choose 3}=56$ possibilities.
If $x_1 = 3$, there are $6$ spaces for the bars and ${6\choose 3} = 20$ possibilities.
Finally, if $x_1 = 5$, there are $4$ spaces for the bars, for a total of ${4\choose 3} = 4$ possibilities.
Hence there are $56+20+4 = 80$ possibilities in total.
Best Answer