Question

If $x$ is positive integer, then $\Delta$ given below
$\left|\begin{array}{ccc}x!& (x+1)!& (x+2)!\\ (x+1)!& (x+2)!& (x+3)!\\ (x+2)!& (x+3)!& (x+4)!\end{array}\right|$ is divisible by

• A. $(x+2)!$
• B. $x!$
• C. $2(x!)(x+1)!(x+2)!$
• D. All of these

Answer 1

The correct option is D All of these

$\Delta =\left|\begin{array}{ccc}x!& (x+1)!& (x+2)!\\ (x+1)!& (x+2)!& (x+3)!\\ (x+2)!& (x+3)!& (x+4)!\end{array}\right|$
Taking $x!$ common from $R_1$, $(x+1)!$ from $R_2$ and $(x+2)!$ from $R_3$, we get
$\Delta =(x!)(x+1)!(x+2)!\left|\begin{array}{ccc}1& x+1& (x+1)(x+2)\\ 1& x+2& (x+2)(x+3)\\ 1& x+3& (x+3)(x+4)\end{array}\right|$
Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Delta =(x!)(x+1)!(x+2)!\left|\begin{array}{ccc}1& x+1& (x+1)(x+2)\\ 0& 1& 2(x+2)\\ 0& 1& 2(x+3)\end{array}\right|=2(x!)(x+1)!(x+2)!$