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Question: \[47C_{4} + \sum^{\underset{5}{r = 1}}{_{}^{52 - r}C}_{3} =\]...

47C4+∑r=1552−rC3=47C_{4} + \sum^{\underset{5}{r = 1}}{_{}^{52 - r}C}_{3} =

A

47⥂C647 ⥂ C_{6}

B

52C552C_{5}

C

15C1515C_{15}

D

None of these

Answer

15C1515C_{15}

Explanation

Solution

47C4+∑r=1552−rC3=51⥂C3+50⥂C3+49⥂C3+48⥂C3+47⥂C3+47⥂C447C_{4} + \sum_{r = 1}^{5}{52 - rC_{3}} =^{51} ⥂ C_{3} +^{50} ⥂ C_{3} +^{49} ⥂ C_{3} +^{48} ⥂ C_{3} +^{47} ⥂ C_{3} +^{47} ⥂ C_{4}

=51⥂C3+50⥂C3+49⥂C3+48⥂C3+48⥂C4=^{51} ⥂ C_{3} +^{50} ⥂ C_{3} +^{49} ⥂ C_{3} +^{48} ⥂ C_{3} +^{48} ⥂ C_{4}

=51⥂C3+50⥂C3+49⥂C3+49⥂C4=^{51} ⥂ C_{3} +^{50} ⥂ C_{3} +^{49} ⥂ C_{3} +^{49} ⥂ C_{4}

x1+x2+.....+x6x_{1} + x_{2} + ..... + x_{6}.