If \(\frac{(n+3)!}{(n+1)!}=110\) then what will be the value of (n)?

Correct answer: 8

Using \((n+3)!=(n+3)(n+2)(n+1)!\),
\[\frac{(n+3)!}{(n+1)!}=(n+3)(n+2).\]
Hence, \((n+3)(n+2)=110\). Since \(10\times11=110\), we get \(n+2=10\), so \(n=8\). The other algebraic root is \(n=-13\), but then \((n+1)!\) is not defined. Exam tip: In a factorial quotient, cancel the common factorial part first and write the remaining consecutive factors.