First, there are $n_0(z-1,d)$ such permutations. On the other hand, we can divide
the them to two types depending on whether $\pi[1]=1$. Those having $\pi[1]\ne 1$
are exactly the $n_0(z,d)$ permutations satisfying~$M_z$. The others correspond to
First, there are $n_0(z-1,d)$ such permutations. On the other hand, we can divide
the them to two types depending on whether $\pi[1]=1$. Those having $\pi[1]\ne 1$
are exactly the $n_0(z,d)$ permutations satisfying~$M_z$. The others correspond to