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The value of $^{47} C_{4} + \sum_{r = 1}^{5}^{(52 - r)} C_{3}$ is equal to

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47C6

52C5

52C4

None of these

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Correct option is C

47C4+∑r=15 52−rC3=47C4+51C3+50C3+49C3+48C3+47C3 =51C3+50C3+49C3+48C3+ 47C3+47C4=52C4

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The value of 47C4+∑r=15 52−rC3 is equal to

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The value of $^{47} C_{4} + \sum_{r = 1}^{5}^{(52 - r)} C_{3}$ is equal to