Problem counting spin states I can't figure out how many different spin states I can create with a four-electron system. I think I can create a spin-zero state, three spin-one states, and five spin-two states. That gives me nine possible states altogether. 
My problem is that I can specify a maximum of eight (complex) numbers to completely describe the spin states of the four electrons. But the nine spin-states I can seemingly create correspond to the spherical harmonic functions and I know for sure that they are linearly independent. It seems very wrong.
This discrepancy in the counting doesn't appear until you get to four electrons. It gets worse as you add more electrons.
Does anyone else have a problem with this?
EDIT: Thanks for the excellent answers, especially from Lagerbeer. It turns out I was even more messed up than I thought I was. I don't run into trouble with four electrons....I'm already in trouble with three. I didn't realize it because I was counting 2n electron-parameters instead of n^2, so I had 2, 4, 6, 8...  (parameters to describe electron spin) as opposed to 2, 4, 8, 16... as people have pointed out. And this has to relate to (l,m) spin states of 2, 4, 6, 9, 12, 16...
So there's already a problem with three electrons and it boils down to this: you have two different states with z-axis spin 1/2: the are (3/2, 1/2) and (1/2, 1/2). To make them up from electrons you have at your disposal these three states:
{A} = duu
{B} = udu
{C} = uud
The obvious thing to do is add the three together (and of course normalize); and I believe when you do, you get the (3/2, 1/2) state. The question is: how do you create the (1/2, 1/2) state? 
I think I know the answer and I've posted it on my blog. Anyone want to take a stab at it?
 A: Ah, this is a very subtle thing, and it's true that it first occurs for four electrons. 
First, here's an easy way to tell how many states you should expect: Just use the "individual-electron's spin-basis". With four electrons, each of them could be up or down spin, so we expect a total of $2^4 = 16$ states. 
So where are you missing the states? Well, there is more than one spin 0 state, for example: You can get a spin 0 state if you combine two electrons each into a spin 0 singlet and then combine those two spin 0 states into an overall spin 0 state. 
But you can also combine two electrons each into a spin 1 state, and then we know from the rules of adding angular momenta that the total angular momentum of two spin 1 systems can be $0$, $1$ or $2$, so you get a second spin-0 state from combining the spin-1 states in a particular way.
So here's the book keeping:
We get two spin 0 states.
We get 9 spin 1 states (3 ways to get a spin 1 state: Either the first pair of electrons has spin 1 and the second spin 0, or the first pair has spin 0 and the second has spin 0, or both have spin 1).
We get 5 spin 2 states.
5 + 9 + 2 = 16.
A: Lagerbaer has already answered OP's question for $n=4$ distinguishable spin dublets. More generally, the number of spin multiplets of $n$ distinguishable spin dublets can be deduced from repeated applications of the $SU(2)$ Clebsch-Gordan fusion rule
$$ \underline{\large\bf 2} \otimes \underline{\large\bf n}~=~  \left\{ \begin{array}{lcl} \underline{\large\bf n+1}~\oplus~\underline{\large\bf n-1} &\text{for}& n\geq 2, \\ \underline{\large\bf n+1}&\text{for}& n=1, \end{array} \right. $$
and the distributive law for $\otimes$ and $\oplus$. Explicitly, the first few tensor powers read
$$  \underline{\large\bf 2}^{\otimes 1}
~=~\underline{\large\bf 2}, $$
$$  \underline{\large\bf 2}^{\otimes 2}
~=~\underline{\large\bf 3}~\oplus~\underline{\large\bf 1} ,$$
$$  \underline{\large\bf 2}^{\otimes 3}
~=~\underline{\large\bf 4}~\oplus~2~\underline{\large\bf 2} ,$$
$$  \underline{\large\bf 2}^{\otimes 4}
~=~\underline{\large\bf 5}~\oplus~3~\underline{\large\bf 3}
~\oplus~2~\underline{\large\bf 1} ,$$
$$  \underline{\large\bf 2}^{\otimes 5}
~=~\underline{\large\bf 6}~\oplus~4~\underline{\large\bf 4}
~\oplus~5~\underline{\large\bf 2} ,$$
$$  \underline{\large\bf 2}^{\otimes 6}
~=~\underline{\large\bf 7}~\oplus~5~\underline{\large\bf 5}
~\oplus~9~\underline{\large\bf 3}~\oplus~5~\underline{\large\bf 1} ,$$
$$ \vdots$$
Here the irreps $\underline{\large\bf 1}$, $\underline{\large\bf 2}$,$\underline{\large\bf 3}$, $\ldots$, denote singlet, dublet, triplet, $\ldots$, i.e., spin $0$, $\frac{1}{2}$, $1$, $\ldots$, respectively. The above pattern resembles Pascal's triangle. Clearly the general formula is of the form
$$  \underline{\large\bf 2}^{\otimes n}~=~\bigoplus_{k=0}^{[\frac{n}{2}]} m_{n,k} ~\underline{\large\bf n+1-2k}, \qquad n\in \mathbb{N}.$$
Here the multiplicities $m_{n,k}\in \mathbb{N}_{0}$ satisfy
$$ m_{n,k}~=~0 \quad\text{for}\quad k> [\frac{n}{2}], $$ 
$$ m_{n,0}~=~1,$$
and
$$ m_{n,k-1}+ m_{n,k}~=~m_{n+1,k}\quad\text{for}\quad k \geq 1.  $$ 
A closed formula for the multiplicities reads (hattip:Trimok)
$$m_{n,k}~=~ \frac{n!~(n + 1 - 2k)}{k!~ (n + 1 - k)!}. $$
A: For four, say, electrons we have 16 ($=2^4$) possible states. 1 With spin +2 and 1 with spin -2, 4 with spin +1 and 4 with spin -1, and at last 6 with spin 0, so in total 16 (the 1,4,6,4,1 correspond to row 4 of the Pascal triangle). See this question for the states of n electrons.
For example, for three electrons we get the state (the aid of the Pascal Triangle obviously pays well of in the case of a large number of electrons):
$$\frac 1 {\sqrt{8}}(+|\uparrow\uparrow\uparrow\rangle+|\uparrow\uparrow\downarrow\rangle+|\uparrow\downarrow\uparrow\rangle+|\downarrow\uparrow\uparrow\rangle-|\uparrow\downarrow\downarrow\rangle-|\downarrow\uparrow\downarrow\rangle-|\downarrow\downarrow\uparrow\rangle-|\downarrow\downarrow\downarrow\rangle)$$
This total wavefunction is normalized and is antisymmetric if we interchange the kets with opposite arrows.
