If \( \frac{(n+4)!}{(n+2)!}=182 \), what is the value of (n)?

Correct answer: 10

Cancelling \((n+2)!\) from \((n+4)!\) gives \(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\). Hence, \((n+4)(n+3)=182\). Since \(182=14\times13\), we get \(n+4=14\), so \(n=10\). For option 9, the product would be \(13\times12=156\), so it is not correct. Exam tip: In a factorial fraction, expand the larger factorial only as far as needed and cancel the common factorial term.