If \( \frac{(n+4)!}{(n-1)!(n+3)(n+2)}=840 \), what is the value of (n)?

Correct answer: No natural-number solution

Using \((n+4)!=(n+4)(n+3)(n+2)(n+1)n(n-1)!\), the expression simplifies to \(n(n+1)(n+4)\). At \(n=7\), it equals \(7\times8\times11=616\), and at \(n=8\), it equals \(8\times9\times12=864\). Since \(n(n+1)(n+4)\) increases for natural numbers, it cannot take the value 840 between these two consecutive cases. Hence, there is no natural-number solution. Exam tip: Expand factorials into consecutive factors first, then cancel common terms.