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6.Permutation and Combination
medium

$\left( {\begin{array}{*{20}{c}}
  {15} \\ 
  {3r} 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {15} \\ 
  {r + 3} 
\end{array}} \right)$  હોય તો $r\,\, = \,\,........$

A

$3$

B

$4$

C

$5$

D

$8$

Solution

$\left( {\begin{array}{*{20}{c}}
  {15} \\ 
  {3r} 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {15} \\ 
  {r + 3} 
\end{array}} \right)$

$\,\left( {\begin{array}{*{20}{c}}
  {15} \\ 
  {15 – 3r} 
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {15} \\ 
  {r + 3} 
\end{array}} \right)\,\,$

$15 – 3r = r + 3$

$\,12 = 4r\,\,$

$r = 3$

Standard 11
Mathematics

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