If \( \frac{(n+3)!}{(n-1)!(n+2)(n+1)}=108 \), what is the value of (n)?

Correct answer: 9

Write \((n+3)!=(n+3)(n+2)(n+1)n(n-1)!\). Cancelling \((n-1)!\), \((n+2)\), and \((n+1)\) gives \(n(n+3)=108\). For \(n=9\), \(9\times12=108\), so the correct answer is 9. For instance, \(n=8\) gives \(8\times11=88\), not 108. Exam tip: In factorial problems, expand the larger factorial up to the smaller one before cancelling common factors.