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Complex Numbers

47 C 4+∑ r =1...

Question

$^{47} C_{4} + 5 \sum r = 1^{52 - r} C_{3}$ =

A

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B

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C

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D

${}^{52}C_3$

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Solution

The correct option is C

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

$=^{51} C_{3} +^{50} C_{3} +^{49} C_{3} +^{48} C_{3} +^{48} C_{4}$

$=^{51} C_{3} +^{50} C_{3} +^{49} C_{3} +^{49} C_{4}$

$=^{51} C_{3} +^{50} C_{3} +^{50} C_{4} +^{51} C_{3} +^{51} C_{4}$ = $^{52} C_{4}$

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