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47C4 + \(\sum_{r=1}^{5}\) \(^{52-r}C_3\) is equal to (a) 47C6 (b) 52C5 (c) 52C4 (d) None of these |
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Answer» Answer :(c) 52C4 \(^{47}C_4 + \sum_{r=1}^5 \)\(^{52-r}C_3\) = 51C3 + 50C3 + 49C3 + 48C3 + 47C3 + 47C4 = 51C3 + 50C3 + 49C3 + 48C3 + 48C4 (\(\because\) nCr + nCr+1 = n+1Cr+1) = 51C3 + 50C3 + 49C3 + 49C4 = 51C3 + 50C3 + 50C4 = 51C3 + 51C4 = 52C4 |
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