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  4. If x is a positive integer, then | x! (x+1)! (x+2)! [0.3em] (x+1)! (x+2)! (x+3)! [0.3em] (x+2)! (x+3)! &(x+4)! | is equal to

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Q. If x is a positive integer, then $\begin{vmatrix} x! & (x+1)! & (x+2)! \\[0.3em] (x+1)! & (x+2)! & (x+3)! \\[0.3em] (x+2)! & (x+3)! &(x+4)! \end{vmatrix}$ is equal to

Determinants

Solution:

$\begin{vmatrix} x! & (x+1)! & (x+2)! \\[0.3em] (x+1)! & (x+2)! & (x+3)! \\[0.3em] (x+2)! & (x+3)! &(x+4)! \end{vmatrix}$
= $x \, ! (x+1) ! (x + 2) !$
$\begin{vmatrix}1&1&1\\ x+1&x+2&\left(x+3\right)\\ \left(x+2\right)\left(x+1\right)&\left(x+2\right)\left(x+3\right)&\left(x+3\right)\left(x+4\right)\end{vmatrix}$
= $2x ! (x + 1 ) ! ( x + 2 ) !$