If \( \frac{(n+6)!}{(n+4)!}+\frac{(n+5)!}{(n+3)!}=338 \), what is the value of (n)?

Correct answer: 8

\(\frac{(n+6)!}{(n+4)!}=(n+6)(n+5)\) and \(\frac{(n+5)!}{(n+3)!}=(n+5)(n+4)\). Thus, their sum is \((n+5)[(n+6)+(n+4)]=2(n+5)^2\). Hence \(2(n+5)^2=338\), so \((n+5)^2=169\). Since the factorial arguments must be valid integers, take \(n+5=13\), which gives \(n=8\). Substituting \(n=9\) gives \(392\), not \(338\). Exam tip: In factorial fractions, cancel the common factorial part first and write the remaining consecutive factors.