What is the simplified form of \( \frac{\frac{(n+3)!}{(n-2)!}}{\frac{(n+2)!}{(n-3)!}} \)?

Correct answer: \(\frac{n+3}{n-2}\)

Writing the complex fraction as a product gives \(\frac{(n+3)!(n-3)!}{(n-2)!(n+2)!}\). Now use \((n+3)!=(n+3)(n+2)!\) and \((n-2)!=(n-2)(n-3)!\). The factors \((n+2)!\) and \((n-3)!\) cancel, leaving \(\frac{n+3}{n-2}\). Option D is the reciprocal of the required result, so it is incorrect. Exam tip: expand a larger factorial only until a common smaller factorial appears, then cancel common factors.