If for some $m,n$, $\binom{6}{m}+2\binom{6}{m+1}+\binom{6}{m+2}>8\binom{6}{3}$ and $\,^{\,n-1}\!P_{3}:\,^{\,n}\!P_{4}=1:8$, then $\,^{\,n}\!P_{\,n+1}+\,^{\,n+1}\!C_{m}$ is equal to: