Problem counting spin states

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.

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)

$$m_{n,k}~=~ \frac{n!~(n + 1 - 2k)}{k!~ (n + 1 - k)!}. $$