Question

Let $f\left(n\right)=\left|\begin{array}{ccc}n& n+1& n+2\\ {}^n{P}_n& {}^{n+1}{P}_{n+1}& {}^{n+2}{P}_{n+2}\\ {}^n{C}_n& {}^{n+1}{C}_{n+1}& {}^{n+2}{C}_{n+2}\end{array}\right|$ then $f\left(n\right)$ is divisible by:

• A. $n^2+n$
• B. $n+1)$
• C. $(n+2)!$
• D. $n!(n^2+n+1)$

Answer 1

The correct option is D $n!(n^2+n+1)$

$\left|\begin{array}{ccc}n& n+1& n+2\\ {}^n{P}_n& {}^{n+1}{P}_{n+1}& {}^{n+2}{P}_{n+2}\\ {}^n{C}_n& {}^{n+1}{C}_{n+1}& {}^{n+2}{C}_{n+2}\end{array}\right|=\left|\begin{array}{ccc}n& n+1& n+2\\ n!& (n+1)!& (n+2)!\\ 1& 1& 1\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ n& n+1& n+2\\ n!& (n+1)!& (n+2)!\end{array}\right|=\left|\begin{array}{ccc}0& 0& 1\\ -1& -1& n+2\\ -n.n!& -(n+1)(n+1)!& (n+2)!\end{array}\right|\begin{array}{c}{C}_1\to C_1-C_2\\ C_2\to C_2-C_3\end{array}=(n+1).(n+1)!-n.n!=n!({(n+1)}^2-n)=n!(n^2+n+1)$