Dimension of spin zero Hilbert subspace Consider a system of 4 different spin-5/2 particles, each transforming in the dimension-6 irreducible representation of $SU(2)$. The dimension of the total Hilbert space is $6^4$. Could anyone show how to compute the dimension of a subspace of total spin zero?
I would like to calculate the number of ways 4 particles combine to spin 0. Is there a general procedure?
 A: As suggested by @Qmechanic, your problem is addressable by inspection,
$$
5/2 \otimes 5/2= 5\oplus 4\oplus 3\oplus 2\oplus 1\oplus 0 , ~~~\leadsto \\
( 5\oplus 4\oplus 3\oplus 2\oplus 1\oplus 0)\otimes  (5\oplus 4\oplus 3\oplus 2\oplus 1\oplus 0).
$$
But you can only get singlets out of the diagonal terms in the distribution of this, 5⊗5, 4⊗4, ...,0⊗0, so, then, six of them. That is, of  your 1296 states, 6 are spinless.

Is there a general procedure?

Yes, eqn (6) & (11) of Curtright, Van Kortryk, and Zachos 2016, with explicit evaluations which have been around forever...(However, admittedly, your eyes will pop out at the $M(0,4;5/2)=146-140$. cf footnote$^\natural$.)
Can you now supplant your 4 with n tending to infinity, should your spirit move you?

$^\natural$
The multiplicity of total spin s in the composition of n spin j multiplets is always given by a difference,
\begin{equation}
M\left(  s;n;j\right)  =c_{0}\left(  s,n,j\right)  -c_{0}\left(
s+1,n,j\right)  \ ,  
\end{equation}
where $2s$ is any integer such that $0\leq2s\leq2nj$, and where $s=0$ is
always allowed when $j$ is an integer but is only allowed for even $n$ when
$j$ is a semi-integer:
\begin{equation} \Large
c_{0}\left(  s,n,j\right)  =\sum_{k=0}^{\left\lfloor \frac{nj+s}
{2j+1}\right\rfloor }\left(-1\right)  ^{k}\binom{n}{k}\binom{nj+s-\left(
2j+1\right)  k+n-1}{nj+s-\left(  2j+1\right)  k}\ .
\end{equation}
