Nếu \(2A_n^4 = 3A_{n - 1}^4\) thì \(n\) bằng:

\(2A_n^4 = 3A_{n - 1}^4\)  \(\left( {n \ge 5;\,\,n \in N} \right)\)

\( \Leftrightarrow 2\dfrac{{n!}}{{\left( {n - 4} \right)!}} = 3\dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 5} \right)!}}\)

\( \Leftrightarrow 2n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) - 3\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right) = 0\)

\( \Leftrightarrow \left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left[ {2n - 3\left( {n - 4} \right)} \right] = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}n = 1\,\,\,\,\left( {ktm} \right)\\n = 2\,\,\,\left( {ktm} \right)\\n = 3\,\,\,\,\left( {ktm} \right)\\2n - 3n + 12 = 0\end{array} \right. \Leftrightarrow n = 12\,\,\left( {tm} \right)\)