ĐK: \(n\ge4\)
\(3A^2_n-2C_{n-2}^2=54\)
\(\Leftrightarrow3\cdot\dfrac{n!}{\left(n-2\right)!}-2\cdot\dfrac{\left(n-2\right)!}{2!\cdot\left(n-4\right)!}=54\)
\(\Leftrightarrow3n\left(n-1\right)-\left(n-3\right)\left(n-2\right)=54\)
\(\Leftrightarrow2n^2+2n-60=0\)
\(\Leftrightarrow\left[{}\begin{matrix}n=5\\n=-6\end{matrix}\right.\)
Do \(n\ge4\) nên \(n=5\)
\(\Leftrightarrow3\cdot\dfrac{n!}{\left(n-2\right)!}-2\cdot\dfrac{\left(n-2\right)!}{\left(n-4\right)!\cdot2!}=54\)
\(\Leftrightarrow3n\left(n-1\right)-\left(n-3\right)\left(n-2\right)=54\)
=>3n^2-3n-n^2+5n-6=54
=>2n^2+2n-6-54=0
=>2n^2+2n-60=0
=>n^2+n-30=0
=>n=5
