Câu hỏi 19

\(\begin{array}{l}C_n^3 = A_n^2 - 10 \Leftrightarrow \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} = \dfrac{{n!}}{{\left( {n - 2} \right)!}} - 10\\ \Leftrightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} = n\left( {n - 1} \right) - 10\\ \Leftrightarrow {n^3} - 3{n^2} + 2n - 6{n^2} + 6n + 60 = 0\\ \Leftrightarrow {n^3} - 9{n^2} + 8n + 60 = 0\\ \Leftrightarrow \left[ \begin{array}{l}n =  - 2\,\,\,\left( {ktm} \right)\\n = 5\\n = 6\end{array} \right.\end{array}\)

Chọn D.