Stars and Bars Question Dealing with Number of Solutions How many solutions are there to the equation $x+y+z = 9$ if $x, y, \text{and }z$ are nonnegative with the restriction $x \le y\le z$?
For this problem I have so far the idea of splitting $x$ into cases. Like if $x=0$, then $y$ ranges from $(0-4) \text{ while $z$ ranges from } z= 9 - y$. But how do I find the number of solutions using this idea?
 A: We know that the total number of solutions to the equation is given by $\binom{11}{9} = 55$. Out of these $55$ solutions, there are certain sets of solutions, in which a number appears more than once.
There are $3$ permutations for the elements of the solution sets having $2$ identical numbers, which are $(0,0,9), (1,1,7), (2,2,5),(1,4,4)$ and there is only $1$ permutation for the set having all $3$ numbers identical, i.e. $(3,3,3)$. i.e. there are a total of $3.4 + 1 = 13$ solutions sets in these $55$ having at least two elements identical. It follows that $55-13 = 42$ solution sets have all elements distinct.
Now, remember that the elements solution set having distinct elements can be permuted in $3! = 6$ ways. But we only want the one case, in which they are arranged in the increasing order. Hence the cases in which the distinct solutions are in increasing order is given by  $42/6 = 7$.
Also, as discussed above, there are also some solution sets having $2$ elements identical, and also all $3$ identical, which are $(0,0,9), (1,1,7), (2,2,5),(1,4,4), (3,3,3)$. We don't want to count their permutations. Hence there are $5$ of these cases.
Hence the total number of solutions satisfying your condition are $\fbox{7+5 = 12}$.
