We have to sum up all \( np-nh \) excitations, and there are \( \binom{n_u}{n} \) particle states and \( \binom{A}{A-n} \) hole states for each \( n \). Thus, we have for the total number
$$ \begin{align} \sum_{n=0}^A \binom{n_u}{n} \binom{A}{A-n}= \binom{A+n_u}{A} . \tag{16} \end{align} $$The right hand side are obviously all ways to distribute \( A \) fermions over \( n_0+A \) orbitals.