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Question

# The value of the determinant $\left|\begin{array}{ccc}15!& 16!& 17!\\ 16!& 17!& 18!\\ 17!& 18!& 19!\end{array}\right|=$

A

$15!+16!$

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B

$2\left(15!\right)\left(16!\right)\left(17!\right)$

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C

$15!+16!+17!$

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D

$16!+17!$

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Solution

## The correct option is B $2\left(15!\right)\left(16!\right)\left(17!\right)$Explanation for correct option $\left|\begin{array}{ccc}15!& 16!& 17!\\ 16!& 17!& 18!\\ 17!& 18!& 19!\end{array}\right|\phantom{\rule{0ex}{0ex}}=\left(15!\right)\left(16!\right)\left(17!\right)\left|\begin{array}{ccc}1& 16& 16·17\\ 1& 17& 17·18\\ 1& 18& 18·19\end{array}\right|\phantom{\rule{0ex}{0ex}}=\left(15!\right)\left(16!\right)\left(17!\right)\left|\begin{array}{ccc}1& 16& 272\\ 1& 17& 306\\ 1& 18& 342\end{array}\right|\phantom{\rule{0ex}{0ex}}=\left(15!\right)\left(16!\right)\left(17!\right)\left|\begin{array}{ccc}0& 2& 70\\ 0& 1& 36\\ 1& 18& 342\end{array}\right|\left[{R}_{2}\to {R}_{3}-{R}_{2},{R}_{1}\to {R}_{3}-{R}_{1}\right]\phantom{\rule{0ex}{0ex}}=\left(15!\right)\left(16!\right)\left(17!\right)\left[1\left(2×36-1×70\right)\right]\phantom{\rule{0ex}{0ex}}=\left(15!\right)\left(16!\right)\left(17!\right)\left[72-70\right]\phantom{\rule{0ex}{0ex}}=\left(15!\right)\left(16!\right)\left(17!\right)2$Hence, the correct option is $\mathrm{Option}\left(\mathrm{B}\right)$

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