Tính \(M = \dfrac{{A_{n + 1}^4 + 3A_n^3}}{{\left( {n + 1} \right)!}}\), biết \(C_{n + 1}^2 + 2C_{n + 2}^2 + 2C_{n + 3}^2 + C_{n + 4}^2 = 149\).

Điều kiện: \(\left\{ \begin{array}{l}n \in \mathbb{N}\\n \ge 3\end{array} \right.\)

Ta có: \(C_{n + 1}^2 + 2C_{n + 2}^2 + 2C_{n + 3}^2 + C_{n + 4}^2 = 149\)

\( \Leftrightarrow \dfrac{{\left( {n + 1} \right)!}}{{2!\left( {n - 1} \right)!}} + 2\dfrac{{\left( {n + 2} \right)!}}{{2!n!}} + 2\dfrac{{\left( {n + 3} \right)!}}{{2!\left( {n + 1} \right)!}} + \dfrac{{\left( {n + 4} \right)!}}{{2!\left( {n + 2} \right)!}} = 149\)

\( \Leftrightarrow \dfrac{{\left( {n + 1} \right)n}}{2} + \dfrac{{2\left( {n + 2} \right)\left( {n + 1} \right)}}{2}\)\( + \dfrac{{2\left( {n + 3} \right)\left( {n + 2} \right)}}{2} + \dfrac{{\left( {n + 4} \right)\left( {n + 3} \right)}}{2} = 149\)

\( \Leftrightarrow {n^2} + n + 2\left( {{n^2} + 3n + 2} \right)\)\( + 2\left( {{n^2} + 5n + 6} \right) + {n^2} + 7n + 12 = 298\)

\( \Leftrightarrow 6{n^2} + 24n - 270 = 0\)\( \Leftrightarrow \left[ \begin{array}{l}n = 5\left( {TM} \right)\\n =  - 9\left( L \right)\end{array} \right.\)

Do đó: \(M = \dfrac{{A_6^4 + 3A_5^3}}{{6!}} = \dfrac{3}{4}\).