Number of solution set such that $a \geq b \geq c \geq d \geq 0$; $a, b, c, d \in \mathbb{Z}$ and $a + b + c + d = 10$ Given $(a, b, c, d)$ is a set of integers and $a \geq b \geq c \geq d \geq 0$. Find the number of solution sets for $a + b + c + d = 10$.
(This is problem from a competition, the answer key says its $23$, wherein my answer is $455$)
Solution
$x_1 + x_2 +1 + x_3 + 2 + x_4 + 3 = 10$
$x_1 + x_2 + x_3 + x_4 = 16$
$_{15}C_3$ = 455

$a \geq b \geq c \geq d \geq 0$ give easily as follows
$$\begin{cases}a=10\text{ gives } 1\\a=9\text{ gives } 1\\a=8\text{ gives }2\\a=7\text{ gives }3\\a=6\text{ gives }4\\a=5\text{ gives }5\\a=4\text{ gives }5\\a=3\text{ gives }2
\end{cases}$$
examples: $8200,8110\text{ are the two solutions for a=8 }\\ 5500,5410,5320,5311,5221\text{ are the five solutions for a=5 }$ 
Thus there are $23$ solutions.
