If \(\frac{(n+2)!}{n!}=42\), what is the value of (n)?

Correct answer: \(5\)

Since \((n+2)!=(n+2)(n+1)n!\), we get \(\frac{(n+2)!}{n!}=(n+2)(n+1)\). Hence, \((n+2)(n+1)=42\). The consecutive positive integers whose product is \(42\) are \(6\) and \(7\), so \(n+1=6\) and \(n=5\). If \(n=4\), the product would be \(5\times6=30\), so it is not correct. Exam tip: Before cancelling a factorial in a fraction, expand the larger factorial up to the common factorial term.