Question

The number of integral solutions to the system of equations $x_1+x_2+x_3+x_4+x_5=20$ and $x_1+x_2=15$ if $x_k\ge 0$ are :

• A. $300$
• B. $350$
• C. $336$
• D. $316$

Answer 1

The correct option is C $336$

Consider the first equation, $x_1+x_2=15$. A way to think of this is that you have 15 objects laid out in a line in front of you, and you can put a divider in any gap, such that the number of balls to the left of the divider will be $x_1$, and those to the right will be $x_2$. The number of available gaps to you is $16$ (because $x_r$ can also be equal to $0$), so the number of ways you can place the divider (i.e., the number of solutions to the equation) is:
${}^{16}{C}_1=16$. (the divider can be placed in any one of 16 gaps)
In general, the number of non-negative integral solutions of the equation $x_1+x_2+x_3+...x_r=n$ is ${}^{n+r-1}{C}_{r-1}$.
A similar logic for the second equation will give the number of solutions as:
${}^7{C}_2=21$ ..... $[x_1+x_2=15⟹x_3+x_4+x_5=5$
So, the total number of solutions, using multiplication principle:
$16\times 21=336$