If \(\frac{(n+4)!}{(n+2)!}=132\), what is the value of (n)?

Correct answer: \(8\)

Cancelling the common factorial terms gives \(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\). Hence, \((n+4)(n+3)=132\). The consecutive factors of 132 are \(11\) and \(12\), so \(n+3=11\) and therefore \(n=8\). If \(n=7\), the product is \(11\times10=110\), so it is not correct. Exam tip: In a factorial fraction, expand only up to the smaller factorial and cancel common terms.