If \( \frac{(n+5)!+(n+4)!}{(n+3)!}=120 \), what is the value of (n)?

Correct answer: 6

Write \((n+5)!=(n+5)(n+4)!\). Then the numerator becomes \((n+4)![(n+5)+1]\), so the expression simplifies to \((n+4)(n+6)=120\). Hence \((n+5)^2=121\), giving \(n=6\) or \(n=-16\). Since \(n+3\) must be a non-negative integer for the factorial to be defined, only \(n=6\) is valid. For instance, option 5 gives a product of 99, not 120. Exam tip: in factorial equations, first express the larger factorial in terms of the smaller one and cancel common factors.