10. Given $2<\frac{A_{n+1}^{5}}{A_{n-1}^{3}} \leqslant 42$. Then the value of $n$ is
$10.4,5,6$.
Since $\frac{A_{n+1}^{5}}{\mathrm{~A}_{n-1}^{3}}=\frac{(n+1) n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)}$ $=n(n+1)$,
thus, $22, \\ n^{2}+n \leqslant 42,\end{array}\right.$ we get
$-7 \leqslant n<-2$ or $1<n \leqslant 6$.
Based on the range of $n$, we have $4 \leqslant n \leqslant 6$ and $n \in$ $\mathbf{N}$, hence $n=4,5,6$.