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sum(k=0)^(6)"^(51-k)C(3) is equal to...

`sum_(k=0)^(6)"^(51-k)C_(3)` is equal to

A

`""^(51)C_(3)-""^(45)C_(3)`

B

`""^(52)C_(4)-""^(45)C_(4)`

C

`""^(52)C_(3)-""^(45)C_(3)`

D

`""^(51)C_(4)-""^(45)C_(4)`

Text Solution

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The correct Answer is:
B
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