Đáp án:
\(39 \le n \le 43\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
C_n^{10} \ge 3C_n^9\\
3C_n^{10} \ge C_n^{11}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 8} \right)\left( {n - 9} \right)}}{{10!}} \ge \dfrac{{3n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 7} \right)\left( {n - 8} \right)}}{{9!}}\\
\dfrac{{3n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 8} \right)\left( {n - 9} \right)}}{{10!}} \ge \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 9} \right)\left( {n - 10} \right)}}{{11!}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 8} \right)\left( {n - 9} \right)}}{{10!}}:\dfrac{{3n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 7} \right)\left( {n - 8} \right)}}{{9!}} \ge 1\\
\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 9} \right)\left( {n - 10} \right)}}{{11!}}:\dfrac{{3n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 8} \right)\left( {n - 9} \right)}}{{10!}} \le 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 8} \right)\left( {n - 9} \right)}}{{1.2.3.4.5.6.7.8.9.10}}.\dfrac{{1.2.3.4.5.6.7.8.9}}{{3n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 7} \right)\left( {n - 8} \right)}} \ge 1\\
\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 9} \right)\left( {n - 10} \right)}}{{1.2.3.4.5.6.7.8.9.10.11}}.\dfrac{{1.2.3.4.5.6.7.8.9.10}}{{3n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)...\left( {n - 8} \right)\left( {n - 9} \right)}} \le 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{n - 9}}{{3.10}} \ge 1\\
\dfrac{{n - 10}}{{11.3}} \le 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
n - 9 \ge 30\\
n - 10 \le 33
\end{array} \right.\\
\to \left\{ \begin{array}{l}
n \ge 39\\
n \le 43
\end{array} \right.\\
\to 39 \le n \le 43
\end{array}\)
