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ગણ $A$ ના સભ્યોની સંખ્યા $2n + 1$ હોય તો ઓછામાં ઓછા $n$ સભ્યો હોય તેવા $A$ ના કેટલા ઉપગણો હશે ?
$2^n$
$2^{2n}$
${2^{2n}}\, + \,\left( {\begin{array}{*{20}{c}}{2n\, + \,1}\\n\end{array}} \right)$
$2^{2n-1}$
Solution
ઉપગણ ની સંખ્યા $ = \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
n
\end{array}} \right)\, + \,\frac{1}{2}\,$
$\left\{ {\left[ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{n\, + \,1}
\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{n\, + \,1}
\end{array}} \right)} \right]\, + \,\left[ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{n\, + \,2}
\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{n\, + \,2}
\end{array}} \right)} \right]} \right.$
$ + \,\left[ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{2n}
\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{2n}
\end{array}} \right)} \right]\, + $
$\left. {\,\left[ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{2n\, + \,1}
\end{array}} \right)\,} \right] + \left[ {\,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{2n\, + \,1}
\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{2n\, + \,1}
\end{array}} \right)} \right]} \right\}$
$ = \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
n
\end{array}} \right)\, + \,\frac{1}{2}\,\left\{ {\left[ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{n\, + \,1}
\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{n\, + \,1}
\end{array}} \right)\,} \right]} \right.\, + $
$\, + \,\left[ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{n\, + \,2}
\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{n\, – \,1}
\end{array}} \right)} \right] + ,\,…,$
$ + \,\left[ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{2n}
\end{array}} \right)\, + \,\,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
1
\end{array}} \right)} \right]\, + \,\left[ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{2n\, + \,1}
\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
0
\end{array}} \right)} \right]$
$\left[ {\because \,\left( {\begin{array}{*{20}{c}}
n \\
r
\end{array}} \right)\, = \,\left( {\begin{array}{*{20}{c}}
n \\
{n\, – \,r}
\end{array}} \right)} \right]$
$ = \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
n
\end{array}} \right)\, + \,\frac{1}{2}\,\left\{ {\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
0
\end{array}} \right)\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
1
\end{array}} \right)\, + \,,\,…,\, + \,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
{2n\, + \,1}
\end{array}} \right)} \right\}$
=$\,\left( {\begin{array}{*{20}{c}}
{2n\, + \,1} \\
n
\end{array}} \right)\, + \,{2^{2n}}$
