Câu hỏi 6

\(C_n^5 < C_n^3\)\(\left( {n \ge 5;\,\,n \in N} \right)\)

\( \Leftrightarrow C_n^5 - C_n^3 < 0\)

\( \Leftrightarrow \dfrac{{n!}}{{5!\left( {n - 5} \right)!}} - \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} < 0\)

\( \Leftrightarrow \dfrac{{\left( {n - 4} \right)\left( {n - 3} \right)\left( {n - 2} \right)\left( {n - 1} \right)n}}{{120}} - \dfrac{{\left( {n - 2} \right)\left( {n - 1} \right)n}}{6} < 0\)

\( \Leftrightarrow \dfrac{{\left( {n - 4} \right)\left( {n - 3} \right)\left( {n - 2} \right)\left( {n - 1} \right)n - 20\left( {n - 2} \right)\left( {n - 1} \right)n}}{{120}} < 0\)

\( \Leftrightarrow \left( {n - 2} \right)\left( {n - 1} \right)n.\left[ {\left( {n - 4} \right)\left( {n - 3} \right) - 20} \right] < 0\)

Mà \(n \ge 5\)\( \Rightarrow \left( {n - 2} \right)\left( {n - 1} \right)n > 0\)

\( \Leftrightarrow \left( {n - 4} \right)\left( {n - 3} \right) - 20 < 0\)

\( \Leftrightarrow {n^2} - 7n + 12 - 20 < 0\)

\( \Leftrightarrow {n^2} - 7n - 8 < 0\)

\( \Leftrightarrow \left\{ \begin{array}{l} - 1 < n < 8\\n \ge 5\end{array} \right. \Rightarrow 5 \le n < 8\)

Chọn C.