If \( \frac{(n+5)!-(n+4)!}{(n+3)!}=169 \), what is the value of (n)?

Correct answer: 9

Using \((n+5)!=(n+5)(n+4)!\), the numerator becomes \((n+4)![(n+5)-1]=(n+4)(n+4)!\). Therefore,
\[
\frac{(n+5)!-(n+4)!}{(n+3)!}
=\frac{(n+4)(n+4)!}{(n+3)!}
=(n+4)^2.
\]
Thus \((n+4)^2=169=13^2\). For all the factorials to be defined, \(n\geq -3\), so \(n+4\) is positive and \(n+4=13\). Hence \(n=9\). Option 8 gives \(12^2=144\), not 169. Exam tip: In factorial expressions, rewrite the larger factorial in terms of the smaller one before cancelling common factors.