Phương pháp giải:

\(A_n^k = \dfrac{{n!}}{{\left( {n - k} \right)!}}\);    \(C_n^k = \dfrac{{n!}}{{\left( {n - k} \right)!k!}}\)

Lời giải chi tiết:

Ta có:

\(\begin{array}{l}C_{n - 1}^4 - C_{n - 1}^3 - \dfrac{5}{4}A_{n - 2}^2 = 0 \Leftrightarrow \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 5} \right)!4!}} - \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 4} \right)!3!}} - \dfrac{5}{4}.\dfrac{{\left( {n - 2} \right)!}}{{\left( {n - 4} \right)!}} = 0\\ \Leftrightarrow \dfrac{{\left( {n - 2} \right)!}}{{\left( {n - 4} \right)!.4!}}.\left( {\left( {n - 1} \right)\left( {n - 4} \right) - 4\left( {n - 1} \right) - \dfrac{5}{4}.4!} \right) = 0 \Leftrightarrow \left( {n - 1} \right)\left( {n - 4} \right) - 4\left( {n - 1} \right) - 30 = 0\\ \Leftrightarrow {n^2} - 5n + 4 - 4n + 4 - 30 = 0 \Leftrightarrow {n^2} - 9n - 22 = 0 \Leftrightarrow \left[ \begin{array}{l}n = 11\\n =  - 2\,(loai)\end{array} \right.\end{array}\)

Vậy, \(n = 11\).

Chọn: B