The number of solutions of the equation $$x_1 + x_2 + x_3 + x_4 = 50$$, where $$x_1 , x_2 , x_3 , x_4$$ are integers with $$x_1 \geq1, x_2 \geq 2, x_3 \geq 0, x_4 \geq 0$$ is

Let there be four values $$a$$, $$b$$, $$c$$, and $$d$$, all greater than or equal to zero, such that $$x_1=a+1$$, satisfying $$x_1\ge 1$$ and, $$x_2= b+2$$, satisfying $$x_2\ge 2$$, and $$x_3=c$$ and $$x_4=d$$.

We have, $$a+1+b+2+c+d=50$$ or $$a+b+c+d=47$$

Since all of the variables are greater than or equal to zero, the number of solutions to this equation will simply be $${}^{47+4-1}C_{4-1} = \dfrac{50*49*48}{6} = 19600$$