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6.Permutation and Combination
easy
If $^{2n}{C_2}{:^n}{C_2} = 9:2$ and $^n{C_r} = 10$, then $r = $
A
$1$
B
$2$
C
$4$
D
$5$
Solution
(b) $\left( {\frac{{(2n)\;!}}{{2\;!\;(2n – 2)\;!}}} \right)\,\,2 = \left( {\frac{{n\;!}}{{2\;!(n – 2)\;!}}} \right)\,\,9$
$ \Rightarrow (2n)(2n – 1)2 = 9n(n – 1) \Rightarrow n = 5$
Now $^5{C_r} = 10 \Rightarrow r = 2$.
Standard 11
Mathematics