If \( \frac{(n+4)!}{(n+2)!}+\frac{(n+3)!}{(n+1)!}=200 \), what is the value of (n)?

Correct answer: 7

\(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)\) and \(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)\). Hence, their sum is \((n+3)[(n+4)+(n+2)]=2(n+3)^2\). Thus \(2(n+3)^2=200\), giving \((n+3)^2=100\). For the valid non-negative integer factorial domain, \(n+3=10\), so \(n=7\). The other algebraic value, \(n=-13\), does not make the factorial terms valid. Exam tip: in factorial quotients, cancel common factors and write the remaining consecutive factors.