xCr xCr+1 xCr+2 yCr yCr+1 yCr+2 zCr zCr+1 zCr+2 is equal to

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xCr x+1Cr+1 x+1Cr+2 yCr y+1Cr+1 y+1Cr+2 zCr z+1Cr+1 z+1Cr+2

xCr x+1Cr+1 x+2Cr+2 yCr y+1Cr+1 y+2Cr+2 zCr z+1Cr+1 z+2Cr+2

xCr xCr+1 x+1Cr+2 yCr yCr+1 y+1Cr+2 zCr zCr+1 z+1Cr+2

xCr x+1Cr+1 xCr+2 yCr y+1Cr+1 yCr+2 zCr z+1Cr+1 zCr+2

answer is A.

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Detailed Solution

Δ= xCr xCr+1 xCr+2 yCr yCr+1 yCr+2 zCr zCr+1 zCr+2 ………… (1)Applying C2→C2+C1, we getΔ= xCr x+1Cr+1 xCr+2 yCr y+1Cr+1 yCr+2 zCr z+1Cr+1 zCr+2.So, (4) is correct option.In (1), applying C3→C3+C2, we getΔ= xCr xCr+1 x+1Cr+2 yCr yCr+1 y+1Cr+2 zCr zCr+1 z+1Cr+2.So, (3) is correct option.In above applying C2→C2+C1, we getΔ= xCr x+1Cr+1 x+1Cr+2 yCr y+1Cr+1 y+1Cr+2 zCr z+1Cr+1 z+1Cr+2So, (1) is correct option.In above applying C3→C3+C2, we getΔ= xCr x+1Cr+1 x+2Cr+2 yCr y+1Cr+1 y+2Cr+2 zCr z+1Cr+1 z+2Cr+2.So, (2) is correct option.

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