Find the number of solutions of the equation $x+y +z +w = 15$ in the following cases:

Question

Find the number of solutions of the equation $x+y +z +w = 15$ in the following cases:

(a) $x,y,z,w \geq 0$

(b) $x,y,z,w > 0$

(c) $x>2, y>-2, z>0, w>-3$

I think I have an idea on how to do this but I'm not entirely sure. I think I have to use binomial coefficients but i would like to know how to apply it correctly.

Answer 1

For $x_{1}+x_{2}+\dots+x_{n}=k$ (where $x_{i}\geq 0$ and $1\leq i\leq n$), there are ${n+k-1}\choose{k}$ integer solutions.

(a) $x,y,z,w \geq 0$

Since $x+y+w+z\geq15$ has the same number of solutions as $x+y+w+z=15$. In this case, $n=4$ and $k=15$. So ${4+15-1 \choose 15}={18 \choose 15}={18 \choose 3}=816.$

(b) $x,y,z,w > 0$

Notice that $x,y,w,z >0$ means $x,y,w,z,$ is at least $1$. So we let $x'=x-1$, $y'=y-1$,$w'=w-1$, and $z'=z-1$, and we have an equation which has the same number of integer solutions as originally posed in (b): $$\begin{align}x'+y'+w'+z'&=15-4\\ &= 11. \\ \end{align}$$ In this case, $n=4$ and $k=11$. So ${4+11-1 \choose 11}={14 \choose 11}={14 \choose 3}=364.$

You probably can do (c) from here.

Answer 2

$a) \text{ }\binom{15+4-1}{4-1}$.

$b) \text{ } x=x'+1, y=y'+1, z=z'+1, w=w'+1 \Rightarrow x'+y'+z'+ w' = 11 \Rightarrow \binom{11+4-1}{4-1}$.

$c) \text{ } x = x'+3, y = y'- 1, z = z'+1, w = w'- 2 \Rightarrow x'+y'+z'+w' = 14 \Rightarrow \binom{14+4-1}{4-1}$.