MCQ
If $n = ^mC_2,$ then the value of $^n{C_2}$ is given by
- ✓$3{(^{m + 1}}{C_4})$
- B$^{m\,\, - \,\,1}{C_4}$
- C$^{m\,\, + \,\,1}{C_4}$
- D$2{(^{m + 2}}{C_4})$
✓
Answer
Correct option: A.
$3{(^{m + 1}}{C_4})$
a
$n{ = ^m}{C_2} = \frac{{m(m - 1)}}{2}$
$n{ = ^m}{C_2} = \frac{{m(m - 1)}}{2}$
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})$
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