Finding Total Number of Electron States with n=3

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For n = 3, the quantum numbers l can be 0, 1, or 2, leading to m values of -l to +l for each l. The calculations show that for l = 2, there are 5 states; for l = 1, there are 3 states; and for l = 0, there is 1 state, totaling 9 states. Each of these states can be occupied by two electrons due to spin, resulting in 18 total electron states. The confusion arises from not accounting for the spin multiplicity correctly in the initial calculation. Understanding the labeling of each independent quantum state clarifies the total count.
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Homework Statement


n = 3
Using the fact that there are two quantum states for each value of l and m because of electron spin. Find the total number of electron states with n = 3.


Homework Equations





The Attempt at a Solution


I've already determined that l = 0, 1, 2
and m = -2, -1, 0, 1, 2

So, given the information I figured it would be 8 * 2 = 16 quantum states.
Unfortunately, it's supposed to be 18 quantum states and I fail to see where they pick up 2 extra ones. Does anyone know what I am doing incorrectly?
 
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Well for n = 3...

Set l = 2:
then m can take on the range of: -2,-1,0,1,2

Set l = 1:
m = {-1,0,1}

l = 0:
m = 0

There's 9 states.

EDIT: Think of, for n = 3, every independent quantum state by labelling your state:

\psi_{n,l,m}: \psi_{3,0,0} \neq \psi_{3,1,-1} \neq \psi_{3,1,0}

and so on..
 
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