= Fermionic state counting
{title2=$\dim\Lambda^N\mathbb C^M=\binom MN$}
Choosing $N$ different complete <one-particle states> from an <orthonormal basis> of size $M$ gives $\binom MN$ independent <Slater determinants>. This is the <binomial coefficient> because only the occupied subset matters: permutations change the sign of the <Slater determinant>, not the physical <quantum state>. Summing over particle numbers gives $2^M$ states in the finite <fermionic Fock space>.
For the diagram's two <spatial orbitals> $A,B$, take energies $0,\Delta$ with $\Delta>0$, each independent of <spin>. In the complete-state order $(A\uparrow,A\downarrow,B\uparrow,B\downarrow)$, the six $N=2$ <occupation number> strings are $1100,1010,1001,0110,0101,0011$. Their total energies are $0,\Delta,\Delta,\Delta,\Delta,2\Delta$, giving energy multiplicities $1,4,1$. For $N=3$, the <ground state> energy is $\Delta$ with two independent <ground states>, according to which <spin orbital> in $B$ is occupied. Across $N=0,1,2,3,4$, the counts are $1,4,6,4,1$, summing to $16$. These are exact counts for this model, not measured atomic energies.
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