If x is positive integer, then Δ given below ∣∣
∣
∣∣x!(x+1)!(x+2)!(x+1)!(x+2)!(x+3)!(x+2)!(x+3)!(x+4)!∣∣
∣
∣∣ is divisible by
A
(x+2)!
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B
x!
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C
2(x!)(x+1)!(x+2)!
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D
All of these
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Solution
The correct option is D All of these Δ=∣∣
∣
∣∣x!(x+1)!(x+2)!(x+1)!(x+2)!(x+3)!(x+2)!(x+3)!(x+4)!∣∣
∣
∣∣ Taking x! common from R1, (x+1)! from R2 and (x+2)! from R3, we get Δ=(x!)(x+1)!(x+2)!∣∣
∣
∣∣1x+1(x+1)(x+2)1x+2(x+2)(x+3)1x+3(x+3)(x+4)∣∣
∣
∣∣ Applying R2→R2−R1,R3→R3−R1 Δ=(x!)(x+1)!(x+2)!∣∣
∣
∣∣1x+1(x+1)(x+2)012(x+2)012(x+3)∣∣
∣
∣∣=2(x!)(x+1)!(x+2)!