Question

$\left|\begin{array}{ccc}1!& 2!& 3!\\ 2!& 3!& 4!\\ 3!& 4!& 5!\end{array}\right|=2016k$, then $k=$_____

• A. $24$
• B. $84$
• C. $\frac{1}{24}$
• D. $\frac{1}{84}$

Answer 1

The correct option is D $\frac{1}{84}$

$\left|\begin{array}{ccc}1!& 2!& 3!\\ 2!& 3!& 4!\\ 3!& 4!& 5!\end{array}\right|$

$=2!3!\left|\begin{array}{ccc}1& 2& 6\\ 1& 3& 12\\ 1& 4& 20\end{array}\right|$

$=2!3!\times 2\left|\begin{array}{ccc}1& 2& 3\\ 1& 3& 6\\ 1& 4& 10\end{array}\right|$

$=24\left|\begin{array}{ccc}1& 2& 3\\ 0& 1& 3\\ 0& 2& 7\end{array}\right|$

$=24(7-6)=24$

Hence, $24=2016k$
$\Rightarrow k=\frac{1}{84}$