If \( \frac{n!}{(n-2)!}=9n \), what is the positive value of (n)?

Correct answer: 10

Since \(n! = n(n-1)(n-2)!\), \(\frac{n!}{(n-2)!}=n(n-1)\). Hence, \(n(n-1)=9n\). For the factorial expression, \(n\geq2\), so dividing by \(n\) gives \(n-1=9\). Therefore, \(n=10\). Option 9 is not correct because it gives \(9\times8=72\) on the left, whereas the right side is \(81\). Exam tip: In factorial quotients, first expand the larger factorial up to the smaller factorial.