જો n = ^mC_2 હોય તો ^n{C_2} મેળવો. | vedclass

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Question

$n{ = ^m}{C_2} = \frac{{m(m - 1)}}{2}$

Since $m$ and $(m-1)$ are two consecutive natural numbers, therefore their product is an even natural number

Vedclass · Accepted Answer

$3{(^{m + 1}}{C_4})$

Vedclass · Answer

$n{ = ^m}{C_2} = \frac{{m(m - 1)}}{2}$

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, $\frac{{m(m - 1)}}{2} = \frac{{{m^2} - m}}{2}$

$	herefore \,\frac{{m(m - 1)}}{2}{C_2} = \frac{{\left( {\frac{{{m^2} - m}}{2}} ight)\left( {\frac{{{m^2} - m}}{2} - 1} ight)}}{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 	imes (m + 1)m(m - 1)(m - 2)}}{{4 	imes 3 	imes 2 	imes 1}}$

$ = 3{(^{m + 1}}{C_4})$

જો $n = ^mC_2$ હોય તો $^n{C_2}$ મેળવો.

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