Question

Let $f\left(x\right)=\left[\begin{array}{ccc}n& n+1& n+2\\ {{}^{n}P}_n& {{}^{n+1}P}_{n+1}& {{}^{n+2}P}_{n+2}\\ {{}^{n}C}_n& {{}^{n+1}C}_{n+1}& {{}^{n+2}C}_{n+2}\end{array}\right]$ where the symbols have their usual meanings. The $|f\left(x\right)|$ is divisible by

• A. $n^2+n+1$
• B. $n!$
• C. Both (a) and (b)
• D. None of these

Answer 1

The correct option is C Both (a) and (b)

$f\left(x\right)=\left[\begin{array}{ccc}n& n+1& n+2\\ {{}^{n}P}_n& {{}^{n+1}P}_{n+1}& {{}^{n+2}P}_{n+2}\\ {{}^{n}C}_n& {{}^{n+1}C}_{n+1}& {{}^{n+2}C}_{n+2}\end{array}\right]$
$=\left|\begin{array}{ccc}n& n+1& n+2\\ n!& (n+1)!& (n+2)!\\ 1& 1& 1\end{array}\right|[\because {{}^{n}P}_n=n!;{{}^{n}C}_n=1]$
Applying $C_2\to C_2-C_1;C_3\to C_3-C_1$
Then, $f\left(x\right)=\left|\begin{array}{ccc}n& 1& 2\\ n!& n.n!& (n^2+3n+1)n!\\ 1& 0& 0\end{array}\right|$
$=\left|\begin{array}{cc}1& 2\\ n.n!& (n^2+3n+1)n!\end{array}\right|=n!(n^2+3n+1)$