If \(\frac{(n+2)!}{n!}=56\) then what is the positive value of (n)?

Correct answer: 6

Since \((n+2)!=(n+2)(n+1)n!\), we have \(\frac{(n+2)!}{n!}=(n+2)(n+1)\). Therefore, \((n+2)(n+1)=56\). The consecutive factors of 56 are \(7\) and \(8\), so \(n+1=7\) and hence \(n=6\). For the close distractor 7, the product is \(9\times8=72\), not 56. Exam tip: In a factorial fraction, expand the larger factorial only up to the smaller factorial and cancel common terms.