Question

How many integral solutions are there to the system of equations $x_1+x_2+x_3+x_4+x_5=20$ and $x_1+x_2=15$ when $x_k\ge 0$?

Answer 1

We have
$x_1+x_2+x_3+x_4+x_5=20...\left(i\right)$
and $x_1+x_2=15...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get two equations
$x_3+x_4+x_5=5...\left(iii\right)$
$x_1+x_2=15...\left(iv\right)$
and given that $x_1\ge 0,x_2\ge 0,x_3\ge 0,x_4\ge 0$ and $x_5\ge 0$
Then the number of solutions of equation $\left(iii\right)$
$={}^{5+3-1}{C}_{3-1}$
$={}^7{C}_2$
$=\frac{7.6}{1.2}=21$
and the number of solutions of equation $\left(iv\right)$
$={}^{15+2-1}{C}_{2-1}$
$={}^{16}{C}_1=16$
Hence the total number of solutions of the given system of equations = $21\times 16=336$