If \( \frac{(n+2)!}{(n-2)!}=5040 \), what is the value of (n)?

Correct answer: 8

Using \((n+2)!=(n+2)(n+1)n(n-1)(n-2)!\), cancel \((n-2)!\) to obtain \((n+2)(n+1)n(n-1)=5040\). Now \(7\times8\times9\times10=5040\), so the four consecutive factors are \(n-1,n,n+1,n+2\) with \(n=8\). The close distractor \(n=9\) gives \(8\times9\times10\times11\), not 5040. Exam tip: expand the larger factorial only down to the smaller factorial before cancelling.