6. permutation and combination — ગણિત ધોરણ 11 સાયન્સ — Question

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})$