If \( \frac{(n+3)!}{(n+1)!}=72 \), what will be the value of (n)?

Correct answer: 6

On cancelling \((n+1)!\) from \((n+3)!\), we get \(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\). Hence, \((n+3)(n+2)=72\). The consecutive factors of 72 are \(8\) and \(9\), so \(n+2=8\) and \(n=6\). For option 7, the product would be \(9\times10=90\), not 72. Exam tip: In a factorial fraction, expand the larger factorial only up to the smaller one and cancel common factors.