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6.Permutation and Combination
medium

$\left( {_{\,4}^{47}} \right) + \sum\limits_{r = 1}^5 {\left( {_{\,\,\,\,3}^{52 - r}} \right)} = .........$

A

$\left( {\begin{array}{*{20}{c}}{47}\\6\end{array}} \right)$

B

$\left( {\begin{array}{*{20}{c}}{52}\\5\end{array}} \right)$

C

$\left( {\begin{array}{*{20}{c}}{52}\\4\end{array}} \right)$

D

$\left( {\begin{array}{*{20}{c}}{52}\\3\end{array}} \right)$

Solution

$\left( {\begin{array}{*{20}{c}}
  {47} \\ 
  4 
\end{array}} \right) + \sum\limits_{r = 1}^5 {\left( {\begin{array}{*{20}{c}}
  {52 – r} \\ 
  3 
\end{array}} \right)} $

$ = \left( {\begin{array}{*{20}{c}}
  {51} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {50} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {49} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {48} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {47} \\ 
  4 
\end{array}} \right)$

$\left( {\begin{array}{*{20}{c}}
  {51} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {50} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {49} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {48} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {48} \\ 
  4 
\end{array}} \right)$

$ = \left( {\begin{array}{*{20}{c}}
  {51} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {50} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {49} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {49} \\ 
  4 
\end{array}} \right)$

$\left( {\begin{array}{*{20}{c}}
  {51} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {50} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {50} \\ 
  4 
\end{array}} \right)$

$ = \left( {\begin{array}{*{20}{c}}
  {51} \\ 
  3 
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {51} \\ 
  4 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {52} \\ 
  4 
\end{array}} \right)$

Standard 11
Mathematics

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