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?

You are missing the fact that a single even value of $z$ represents several different soluitons, not one.
With $z$ equal to 10, there is only one possible solution (0+0+10).
With $z$ equal to 8, there are 3 possible solutions (0+2+8, 1+1+8, 2+0+8).
With $z$ equal to 6, there are 5 possible solutions.
With $z$ equal to 4, there are 7 possible solutions.
With $z$ equal to 2, there are 9 possible solutions.
With $z$ equal to 0, there are 11 possible solutions.
In total you have 36 solutions with $z$ being even and 66 solutions in total. So the probability is 36/66 or 6/11.
