left( {_{,1}^{10}} right) + left( {_{,2}^{10} | vedclass

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Question

$\left( {\begin{array}{*{20}{c}}
  n \ 
  r 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  n \

Vedclass · Accepted Answer

$\left( {_{\,5}^{14}} \right)$

Vedclass · Answer

$\left( {\begin{array}{*{20}{c}}
  n \ 
  r 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  n \ 
  {r - 1} 
\end{array}} ight) = \left( {\begin{array}{*{20}{c}}
  {n + 1} \ 
  r 
\end{array}} ight)\,\,$

$\left( {\begin{array}{*{20}{c}}
  {10} \ 
  1 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {10} \ 
  2 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {11} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {12} \ 
  4 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {13} \ 
  5 
\end{array}} ight)$

$ = \left( {\begin{array}{*{20}{c}}
  {11} \ 
  2 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {11} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {12} \ 
  4 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {13} \ 
  5 
\end{array}} ight)\,\,$

$ = \left( {\begin{array}{*{20}{c}}
  {12} \ 
  3 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {12} \ 
  4 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {13} \ 
  5 
\end{array}} ight) = \left( {\begin{array}{*{20}{c}}
  {13} \ 
  4 
\end{array}} ight) + \left( {\begin{array}{*{20}{c}}
  {13} \ 
  5 
\end{array}} ight)$

$ = \left( {\begin{array}{*{20}{c}}
  {14} \ 
  5 
\end{array}} ight)$ મળે

$\left( {_{\,1}^{10}} \right) + \left( {_{\,2}^{10}} \right) + \left( {_{\,3}^{11}} \right) + \left( {_{\,4}^{12}} \right) + \left( {_{\,5}^{13}} \right) = ...........$

Similar Questions

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