If \( \frac{(2n)!}{(2n-3)!}=720 \), what is the value of (n)?

Correct answer: 5

Cancelling \((2n-3)!\) from \((2n)!\) gives \(\frac{(2n)!}{(2n-3)!}=(2n)(2n-1)(2n-2)\). Hence, \((2n)(2n-1)(2n-2)=720\). Since \(10\times9\times8=720\), we get \(2n=10\), so \(n=5\). For option 6, the product would be \(12\times11\times10\), not 720. Exam tip: In factorial ratios, expand the larger factorial only up to the smaller factorial and then cancel common terms.