If \(\frac{n!}{(n-1)!}+2=9\), what is the value of (n)?

Correct answer: \(7\)

Since \(n!=n\times(n-1)!\), we get \(\frac{n!}{(n-1)!}=n\) (for \(n\geq1\)). Thus, the given equation becomes \(n+2=9\), which gives \(n=7\). Option \(8\) is a close distractor, but it makes the left-hand side \(8+2=10\), not 9. Exam tip: In a fraction involving consecutive factorials, expand the larger factorial in terms of the smaller one before simplifying.