= Solution
The <Pauli exclusion principle> allows each one-particle magnetic substate only once, since the <fermions> share the other <quantum numbers>. To maximize the total <angular momentum> projection, occupy $m_j=5/2,3/2,1/2,-1/2$, giving
$$
M_{\max}=\frac52+\frac32+\frac12-\frac12=4.
$$
The corresponding <Slater determinant> is a <highest-weight vector>: every attempted raising operation either tries to exceed $j$ or duplicates an occupied substate and vanishes. Therefore it belongs to a multiplet with \b[$\boxed{J_{\max}=4}$], not the distinguishable-particle value $4j=10$. Equivalently, four particles in a six-state shell have two holes whose maximum <angular momentum> projection is $5/2+3/2=4$. The general <maximum total angular momentum of equivalent fermions> is $J_{\max}=Nj-N(N-1)/2$ for $N\leq2j+1$.
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