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

Correct answer: 6

The given expression can be written as \(\frac{(n+2)!}{(n-2)!}\times\frac{(n-3)!}{(n+1)!}\). Cancelling common factorial factors gives \(\frac{n+2}{n-2}\). Hence, \(\frac{n+2}{n-2}=2\Rightarrow n+2=2n-4\Rightarrow n=6\). Therefore, 6 is correct. If 5 is substituted, the value is \(\frac{7}{3}\), not 2. Exam tip: In factorial fractions, expand only enough consecutive factors to cancel common terms.