Given the list of six positive integers 9, 13, $a$, 12, $b$, and 8, how many ordered pairs $(a,b)$ will result in a mean of 10 for the entire list? Given the list of six positive integers $9$, $13$, $a$, $12$, $b$, and $8$, how many ordered pairs $(a,b)$ will result in a mean of $10$ for the entire list?
I took the sum of these six numbers divided by 6 and set it equal to 10. Then I got the equation $a+b=18$. I found there are $17$ pairs that satisfy these conditions, examples $(1,17),(2,16),(3,15),(4,14),\dots,(15,3),(16,2),(17,1)$. Am I missing anything or does this make sense?

We want $\frac{8+9+12+13+a+b}{6} = 10$. We have $8+9+12+13 = 42$, so we are looking for two positive integers $a$ and $b$ that sum to $60 - 42 = 18$. Because we are looking at ordered pairs, $a = 1$, $b=17$ is a distinct solution from $a = 17$, $b= 1$. There are $17$ possible values of $a$ for which there is a corresponding value of $b$ (that is, $a$ can equal any integer from $1$ to $17$, inclusive), so our final answer is $\boxed{17}$.
