Question

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

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

Answer 1

Answer: (A), (C)

n!

$f\left(n\right)=\left|\begin{array}{ccc}n& n+1& n+2\\ {}^n{P}_n& {}^{n+1}{P}_{n+1}& {}^{n+2}{P}_{n+2}\\ {}^n{C}_n& {}^{n+1}{C}_{n+1}& {}^{n+2}{C}_{n+2}\end{array}\right|=n!\left|\begin{array}{ccc}n& (n+1)& (n+2)\\ 1& (n+1)& (n+1)(n+2)\\ 1& 1& 1\end{array}\right|\Delta =n!\left|\begin{array}{ccc}n& 1& 1\\ 1& n& (n+1{)}^2\\ 1& 0& 0\end{array}\right|(C_3\to C_3-C_2,C_2\to C_2-C_1)\Delta =n!\times \left[\right(n+1{)}^2-n]=n!\times (n^2+n+1)$