MCQ
If $^{10}{C_r}{ = ^{10}}{C_{r + 2}}$, then $^5{C_r}$ equals
- A$120$
- B$10$
- C$360$
- ✓$5$
$\Rightarrow r + r + 2 = 10 $
$\Rightarrow r = 4$
$\therefore $$^5{C_r}{ = ^5}{C_4} = \frac{{5\;!}}{{1\;!\;4\;!}} = 5$.
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