Question

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

• A. $n^2+n+1$
• B. $(n+1)!$
• C. $n!$
• D. none of these

Answer 1

Answer: (A), (C)

$n^2+n+1$
$n!$

$\left|\begin{array}{ccc}n& n+1& n+2\\ {}^n{P}_n& {}^{n+1}{P}_{n+1}& {}^{n+2}{P}_{n+2}\\ {}^n{C}_n& {}^{n+1}{C}_{n+1}& {}^{n+2}{C}_{n+2}\end{array}\right|,$
$=\left|\begin{array}{ccc}n& n+1& n+2\\ n!& (n+1)!& (n+2)!\\ 1& 1& 1\end{array}\right|$

Doing,$C_1=C_1-C_2$ and $C_2=C_2-C_3$,

$=n!\left|\begin{array}{ccc}-1& -1& n+2\\ -n& -n^2-2n-1& (n+2)(n+1)\\ 0& 0& 1\end{array}\right|$

$=-1(-n^2-2n-n-0)+(-n-0)+\left(0\right)$

$=n!(n^2+2n+1-n)$

$=n!(n^2+n+1)$

So, it is divisible by $n!$ and $(n^2+n+1)$