If \( \frac{(n+5)!}{(n+3)!}=132 \), what is the value of (n)?

Correct answer: 7

On cancelling \((n+3)!\) from \(\frac{(n+5)!}{(n+3)!}\), we get \((n+5)(n+4)=132\). The consecutive factors of 132 are \(11\) and \(12\), so \(n+4=11\) and hence \(n=7\). For instance, if \(n=6\), the product is \(11\times10=110\), so it is not correct. Exam tip: In factorial quotients, expand the larger factorial only up to the smaller factorial and cancel common terms.