= Solution
The input squared norm is $(N!)^2$, so divide its stated expression by $N!$. Commutativity and the inverse channel transformation give the normalized output
$$
|\psi_{\rm out}\rangle=
\frac1{2^NN!}\left[(b_1^\dagger)^2-(b_2^\dagger)^2\right]^N|0\rangle.
$$
After the binomial expansion, the coefficient of $|2j,2N-2j\rangle$ is
$$
c_j=\frac{(-1)^{N-j}}{2^NN!}\binom Nj
\sqrt{(2j)!(2N-2j)!},\qquad 0\le j\le N.
$$
Hence every possible occupancy has even particle number in each port:
$$
\boxed{P(2j,2N-2j)=\frac1{4^N}\binom{2j}{j}
\binom{2N-2j}{N-j},\quad j=0,\ldots,N.}
$$
All other occupancies have zero <probability>. The normalization identity
$$
\sum_{j=0}^N\binom{2j}{j}\binom{2N-2j}{N-j}=4^N
$$
follows by taking the coefficient of $z^N$ in $(1-4z)^{-1/2}(1-4z)^{-1/2}=(1-4z)^{-1}$. For $N=1$ it reduces to (c), and <symmetry> under $j\leftrightarrow N-j$ gives equal mean occupancies in both outputs.
\Image[/past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2015/paper-1-boson-occupancies.png]
{title=Original balanced-splitter occupations for two-particle statistics and twin-Fock bosons, with independent distinguishable-particle comparisons}
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