More on valence electrons

For this question, how come we can't reach the right answer if we assume l = 0 and 1 since the angular momentum number is = 2, and therefore n would =3 since l = n-1.  Then using 2n^2 to find the number of electrons, we would get 2 x 9=18.  But this isn't the right answer, what are we not accounting for when we approach the problem in this way?
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The problem with this approach is that saying the angular momentum quantum number is 2, and that the principal quantum number is 3, are not the same thing.  Within principle quantum number 3, we can have eighteen electrons:

n = 3
l = 0, 1, 2
m(l) = -2, -1, 0, 1, 2
m(s) = +/- 1/2

This gives us eighteen possible combos.  Each are in the form n, l, m(l), m(s):

s electrons:
3, 0, 0, +1/2
3, 0, 0, -1/2

p electrons:
3, 1, -1, +1/2
3, 1, -1, -1/23, 1, 0, +1/2
3, 1, 0, -1/2
3, 1, 1, +1/2
3, 1, 1, -1/2

d electrons:
3, 2, -2, +1/2
3, 2, -2, -1/23, 2, -1, +1/2
3, 2, -1, -1/2
3, 2, 0, +1/2
3, 2, 0, -1/23, 2, 1, +1/2
3, 2, 1, -1/23, 2, 2, +1/2
3, 2, 2, -1/2

But how many of these actually have l = 2?  Only the last ten.  Thus, there are ten electrons with l = 2.  This question can also be made easier by just remembering that l = 2 refers to d-block electrons.  How many d electrons are there?  10!