We have $x_1 + x_2 + x_3 + x_4 + x_5 = 21 (0\leq x_1\leq 3,1\leq x_2<4,x_3\geq 15)$. Note that $x_2$ is strictly less than 4. Hence to help us solve the equation, we modify it by subtracting the lower bounds of the constraints from the total sum i.e 21. By doing that, there is also a modification to the upper bounds due to the fact that $x_2$ and $x_3$ are already assigned some value i.e 1 and 15 respectively.

We then have $x_1 + x_2' + x_3' + x_4 + x_5 = 5 (0\leq x_1\leq 3,0\leq x_2'\leq 2,x_3'\geq 0)$. The number of solutions for this equation without considering upper bounds is $\binom{5+5-1}{5} = \binom{9}{5} = 126$. Now we can subtract the #solutions for $x_1\geq4$ and $x_2'\geq3$. Both these conditions cannot hold simultaneously since the sum should be 5 and thus we needn't consider the intersection of the cases. We then would obtain $x_1 + x_2' + x_3' + x_4 + x_5 = 1$ and $x_1 + x_2' + x_3' + x_4 + x_5 = 2$ giving us 5 and 15 as #solutions respectively. (P.S: The second highlighted portion arises from solving for $x_2'\geq3$). Subtracting these figures from 126 would fetch us 106.