# How many solutions are there to the equation $x + y + z + w = 17$?

How many solutions are there to the equation $x + y + z + w = 17$?

I don't know if I'm doing this right, but I guessed that the solution would be $\binom{20}{3}$, which equals $1140$. Am I doing this right?

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Are $x, y, z, w$ meant to take positive integer values? Non-negative integer values? Something else? –  Srivatsan Dec 15 '11 at 12:33
They are non-negative integers –  user754950 Dec 15 '11 at 12:38
See math.stackexchange.com/questions/86820/… and the questions referenced there. –  Gerry Myerson Dec 15 '11 at 12:38

Your result is correct. You are looking for the number of monomials of total degree $17$ in $4$ variables. There are $$\binom{17+4-1}{17}=\binom{20}{17}=\frac{20\cdot 19\cdot 18}{3\cdot 2} = 20\cdot19\cdot3 = 1140$$ of these.

A proof would be the following. We associate to any solution $(x,y,z,w)$ a weakly increasing sequence of length $17$, where $0$ appears $x$ times, $1$ appears $y$ times, and so on. For instance, the solution $(4,3,2,8)$ would correspond to $$(0,0,0,0,1,1,1,2,2,3,3,3,3,3,3,3,3)$$ By adding $i$ to the $i$-th component of that sequence, it becomes strictly increasing. This way, we can see that the solutions correspond to subsets of $\{1,\ldots,17+4-1\}$ of cardinality $17$. Of course, you can replace $17$ and $4$ by $n$ and $d$ to obtain a more general statement.

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Line up $20$ skittles in a row:

+ + + + + + + + + + + + + + + + + + + +


Knock any three of them over:

+ + + x + + + + + + x x + + + + + + + +


Now let $x,y,z,$ and $w$ be the number of skittles left standing between the knocked-over skittles. In this case $x=3, y=6, z=0,$ and $w=8$.