MCQ
If $\alpha { = ^m}{C_2}$, then $^\alpha {C_2}$is equal to
- A$^{m + 1}{C_4}$
- B$^{m - 1}{C_4}$
- C$3\,.{\;^{m + 2}}{C_4}$
- ✓$3\;.{\;^{m + 1}}{C_4}$
✓
Answer
Correct option: D.
$3\;.{\;^{m + 1}}{C_4}$
d
(d) $\alpha { = ^m}{C_2} \Rightarrow \alpha = \frac{{m(m - 1)}}{2}$
(d) $\alpha { = ^m}{C_2} \Rightarrow \alpha = \frac{{m(m - 1)}}{2}$
$\therefore $$^\alpha {C_2}{ = ^{m(m - 1)/2}}{C_2} = \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 + 1)$
$ = \frac{1}{8}(m + 1)\;m(m - 1)(m - 2) = 3\;.{\;^{m + 1}}{C_4}$
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