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જો $n = ^mC_2$ હોય તો $^n{C_2}$ મેળવો.
$3{(^{m + 1}}{C_4})$
$^{m\,\, - \,\,1}{C_4}$
$^{m\,\, + \,\,1}{C_4}$
$2{(^{m + 2}}{C_4})$
Solution
$n{ = ^m}{C_2} = \frac{{m(m – 1)}}{2}$
Since $m$ and $(m-1)$ are two consecutive natural numbers, therefore their product is an even natural number. So $\frac{{m(m – 1)}}{2}$ is also a natural number.
Now, $\frac{{m(m – 1)}}{2} = \frac{{{m^2} – m}}{2}$
$\therefore \,\frac{{m(m – 1)}}{2}{C_2} = \frac{{\left( {\frac{{{m^2} – m}}{2}} \right)\left( {\frac{{{m^2} – m}}{2} – 1} \right)}}{2}$
$ = \frac{{m(m – 1)({m^2} – m – 2)}}{8}$
$ = \frac{{m(m – 1)[{m^2} – 2m + m – 2]}}{8}$
$ = \frac{{m(m – 1)[m(m – 2) + 1(m – 2)]}}{8}$
$ = \frac{{m(m – 1)(m – 2)(m + 1)}}{8}$
$ = \frac{{3 \times (m + 1)m(m – 1)(m – 2)}}{{4 \times 3 \times 2 \times 1}}$
$ = 3{(^{m + 1}}{C_4})$
