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જો $\alpha = \left( {\begin{array}{*{20}{c}}
m \\
2
\end{array}} \right)\,\,$ હોય ,તો $\left( {\begin{array}{*{20}{c}}
\alpha \\
2
\end{array}} \right) = ......$
$\left( {\,_{\,\,\,\,4}^{m + 1}\,} \right)$
$\left( {\,_{\,\,\,\,4}^{m - 1}\,} \right)$
$3 \cdot \left( {\,_{\,\,\,\,\,4}^{m + 2}\,} \right)$
$3 \cdot \left( {\,_{\,\,\,\,\,4}^{m + 1}\,} \right)$
Solution
$\alpha = \left( {\begin{array}{*{20}{c}}
m \\
2
\end{array}} \right)\,$ $\alpha = \frac{{m(m – 1)}}{2}\,$
હવે $\,\left( {\begin{array}{*{20}{c}}
\alpha \\
2
\end{array}} \right)\,\,\, = \left( {\frac{{m(m – 1)}}{{\begin{array}{*{20}{c}}
2 \\
2
\end{array}}}} \right)\,$
$ = \frac{1}{2}.\frac{{m(m – 1)}}{2}\,.\,\left[ {\frac{{m(m – 1)}}{2} – 1} \right] = \frac{1}{8}m(m – 1)({m^2} – m – 2)$
$ = \frac{1}{8}m(m – 1)(m + 1)(m – 2)$
$ = 3\left[ {\frac{1}{{24}}(m + 1)(m)(m – 1)(m – 2)} \right]$
$ = 3.\frac{{(m + 1)(m)(m – 1)(m – 2)}}{{4\,!}} = 3\,.\,\left( {\begin{array}{*{20}{c}}
{m + 1} \\
4
\end{array}} \right)$
