Say that your first draw is x and the second draw is y. If you want x+y = 12, you can't have x = 12,13 or 14. In fact, since y is at least one, x+y \ge x+1 > x
So, assume that the first draw is x \in {1, 2, ..., 11}. How many "good" values for y we have for each of these draws?
Well, if x=1, we must draw y = 11 in order to have x+y=12. If x=2, y must be 10, and so on. Since we allow replacement, we can include the case x=y=6 as well.
So, we have 11 possible values for x, each yielding exactly one value for y in order to have x+y=12.
It is actually easy to enumerate all the possible ways:
x = 1 and y = 11
x = 2 and y = 10
x = 3 and y = 9
x = 4 and y = 8
x = 5 and y = 7
x = 6 and y = 6
x = 7 and y = 5
x = 8 and y = 4
x = 9 and y = 3
x = 10 and y = 2
x = 11 and y = 1
