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

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

A
$^{47} C_{6}$
B
$^{52} C_{5}$
C
$^{52} C_{4}$
D
None of these

Solution:

$^{47} C_{4} + \sum_{r = 1}^{5}^{52 - r} C_{3} =^{47} C_{4} +^{51} C_{3} +^{50} C_{3} +^{49} C_{3} +^{48} C_{3} +^{47} C_{3}$

$=^{51} C_{3} +^{50} C_{3} +^{49} C_{3} +^{48} C_{3} + (^{47} C_{3} +^{47} C_{4}) =^{52} C_{4}$

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