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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