If \(\frac{n!}{(n-2)!}=56\), what is the value of (n)?

Correct answer: \(8\)

Using the factorial definition, \(n!=n(n-1)(n-2)!\). Therefore, \(\frac{n!}{(n-2)!}=n(n-1)=56\). Since \(8\times7=56\), \(n=8\). The close distractor \(7\) gives \(7\times6=42\), not 56. In factorial notation, \(n\) is taken as a positive integer, so a negative algebraic root is not valid. Exam tip: Expand the larger factorial only up to the smaller factorial before cancelling.