Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question

Three randomly chosen non negative integers $x, y  \text{  and  } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?

My Approach
Calculating Sample space -:
Number of possible solution for $x + y + z = 10$
$$=\binom{10+3-1}{10}=12 \times 3=66$$
Possible outcome for $z$ to be even =$6(0,2,4,6,8,10)$
Hence the required probability$$=\frac{6}{66}=\frac{1}{11}$$
But the answer is $\frac{6}{11}$
Am I missing something?

This is in fact a composition problem. We want to find all compositions of $x$ and $y$ such that $z$ takes values in $0,2,4,6,8,10$.
Then we count All 2-Compositions of $0,2,4,6,8,10$, that is how can we add 2 numbers to get these values. 
Let $C(n,k,a,b)$ be all k-compositions of n using the numbers from a to b.
We want:
$\sum_{i \in (0,2,4,6,8,10)} C(i,2,0,i)$ = 1+3+5+7+9+11=36.
We know $C(n,k,0,n)=$ $\sum_{i=2}^{10} [{i-1 \choose 1}+2]+1=66$ since we need 2 extra to account for 0 and 1 at the end to account for $x=0$ and $y=0$.
So we get $\frac {36}{66}$ as required.
