If ^{2n}{C_2}{:^n}{C_2} = 9:2 and ^n{C_r} = 1 | vedclass

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Question

(b) $\left( {\frac{{(2n)\;!}}{{2\;!\;(2n - 2)\;!}}} \right)\,\,2 = \left( {\frac{{n\;!}}{{2\;!(n - 2)\;!}}} \right)\,\,9$

$ \Rightarrow (2n)(2n - 1

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(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 $^5{C_r} = 10 \Rightarrow r = 2$.

If $^{2n}{C_2}{:^n}{C_2} = 9:2$ and $^n{C_r} = 10$, then $r = $

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