InterviewSolution
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InterviewSolution
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`|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z)C_(r+2)):}|` is equal to |
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Answer» Correct Answer - A::B::C::D `Delta =|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z)C_(r+2)):}|` Applying `C_(2) to C_(2) +C_(1)` we get `Delta = |{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z)C_(r+2)):}|` so (4) is correct option In (1) applying `C_(3) to C_(3) +C_(2)` we get `Delta =|{:(.^(x)C_(r),,.^(x)C_(r+1),,.^(x+1)C_(r+2)),(.^(y)C_(r),,.^(y)C_(r+1),,.^(y+1)C_(r+2)),(.^(z)C_(r),,.^(z)C_(r+1),,.^(z+1)C_(r+2)):}|` So (3) is correct option In above applying `C_(2) to C_(2) +C_(1)` we get `Delta =|{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x+1)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y+1)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z+1)C_(r+2)):}|` So (1) is correct option In above applying `C_(3) to C_(3)+C_(2)` we get `Delta = |{:(.^(x)C_(r),,.^(x+1)C_(r+1),,.^(x+2)C_(r+2)),(.^(y)C_(r),,.^(y+1)C_(r+1),,.^(y+2)C_(r+2)),(.^(z)C_(r),,.^(z+1)C_(r+1),,.^(z+2)C_(r+2)):}|` so (2) is correct option |
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