If \(\frac{(n+3)!}{(n+1)!}=56\), what is the value of (n)?

Correct answer: 5

Write \((n+3)!=(n+3)(n+2)(n+1)!\). Cancelling \((n+1)!\) gives \((n+3)(n+2)=56\). The consecutive factors of 56 are 7 and 8, so \(n+2=7\) and hence \(n=5\). For the close distractor 4, \((4+3)(4+2)=7\times6=42\), not 56. Exam tip: In a factorial fraction, expand the larger factorial only as far as the smaller factorial, then cancel common factors.