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(B) Given $\Delta = \left| \begin{array}{ccc} x! & (x+1)! & (x+2)! \ (x+1)! & (x+2)! & (x+3)! \ (x+2)! & (x+3)! & (x+4)! \end{array} ight|$.Taking

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$2(x!)(x+1)!(x+2)!$

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(B) Given $\Delta = \left| \begin{array}{ccc} x! & (x+1)! & (x+2)! \ (x+1)! & (x+2)! & (x+3)! \ (x+2)! & (x+3)! & (x+4)! \end{array} ight|$.Taking $x!$ common from $C_1$,$(x+1)!$ common from $C_2$,and $(x+2)!$ common from $C_3$:$\Delta = x!(x+1)!(x+2)! \left| \begin{array}{ccc} 1 & (x+1) & (x+2)(x+1) \ 1 & (x+2) & (x+3)(x+2) \ 1 & (x+3) & (x+4)(x+3) \end{array} ight|$.Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_2$:$\Delta = x!(x+1)!(x+2)! \left| \begin{array}{ccc} 1 & x+1 & (x+2)(x+1) \ 0 & 1 & (x+2)(x+3-x-1) \ 0 & 1 & (x+3)(x+4-x-2) \end{array} ight|$$= x!(x+1)!(x+2)! \left| \begin{array}{ccc} 1 & x+1 & (x+2)(x+1) \ 0 & 1 & 2(x+2) \ 0 & 1 & 2(x+3) \end{array} ight|$.Expanding along $C_1$:$\Delta = x!(x+1)!(x+2)! [1 \cdot (2(x+3) - 2(x+2))]$$= x!(x+1)!(x+2)! [2x + 6 - 2x - 4]$$= x!(x+1)!(x+2)! [2]$$= 2(x!)(x+1)!(x+2)!$.

If $x$ is a positive integer,then $\Delta = \left| \begin{array}{ccc} x! & (x+1)! & (x+2)! \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! \end{array} \right|$ is equal to

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