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The value of $^{50} C_{4} + \sum_{r = 1}^{6}^{56 - r} C_{3}$ is

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^{55} C_{4}

^{55} C_{3}

^{56} C_{3}

^{56} C_{4}

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detailed solution

Correct option is D

^{50} C_{4} + \sum_{r = 1}^{6}^{56 - r} C_{3} =^{50} C_{4} +^{55} C_{3} +^{54} C_{3} +^{53} C_{3} +^{52} C_{3} +^{51} C_{3} +^{50} C_{3}

$= (^{50} C_{4} +^{50} C_{3}) +^{51} C_{3} +^{52} C_{3} +^{53} C_{3} +^{54} C_{3} +^{55} C_{3}$

$= (^{51} C_{4} +^{51} C_{3}) +^{52} C_{3} +^{53} C_{3} +^{54} C_{3} +^{55} C_{3}$

On continuing the process, we get $^{56} C_{4}$

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The value of 50C4+∑r=16 56−rC3 is

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The value of $^{50} C_{4} + \sum_{r = 1}^{6}^{56 - r} C_{3}$ is