If x is a positive integer, and ∆(x) $$=$$ x!(x+1)!(x+2)!(x+1)!(x+2)!(x+3)!(x+2)!(x+3)!(x+4)!, then ∆(x) is equal to

Correct option is 22x!(x+1)! (x+2)!

Questions

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

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2x!(x+1)!

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

2x! (x+3)!

2(x+1)! (x+2)! (x+3)!

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detailed solution

Correct option is B

Taking x! common from R1,(x+1)! from R2 and (x+2)! from R3 to obtainΔ(x)=x!(x+1)!(x+2)!Δ1(x) whereΔ1(x)=1 x+1 (x+1)(x+2)1 x+2 (x+2)(x+3)1 x+3 (x+3)(x+4)Applying R3→R3−R2,R2→R2−R1, we getΔ1(x)=1x+1(x+1)(x+2)012(x+2)012(x+3)=2

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If x is a positive integer, and ∆(x) = x!(x+1)!(x+2)!(x+1)!(x+2)!(x+3)!(x+2)!(x+3)!(x+4)!, then ∆(x) is equal to

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