If ^{2n}{C_3}:{,^n}{C_2} = 44:3, then for whi | vedclass

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Question

(b) $\frac{{(2n)\;!}}{{(2n - 3)\;!\;.\;3\;!}} 	imes \frac{{2\;!\; 	imes (n - 2)\;!}}{{n\;!}}\; = \frac{{44}}{3}$

$ \Rightarrow 4(2n - 1) = 44$ $

Vedclass · Accepted Answer

$r = 4$

Vedclass · Answer

(b) $\frac{{(2n)\;!}}{{(2n - 3)\;!\;.\;3\;!}} 	imes \frac{{2\;!\; 	imes (n - 2)\;!}}{{n\;!}}\; = \frac{{44}}{3}$

$ \Rightarrow 4(2n - 1) = 44$ $ \Rightarrow $ $2n = 12$ $ \Rightarrow $ $n = 6$

Now $^6{C_r} = 15$ $ \Rightarrow $$^6{C_r}{ = ^6}{C_2}$ or $^6{C_4}$ $ \Rightarrow $$r = 2,\;4$.

If $^{2n}{C_3}:{\,^n}{C_2} = 44:3$, then for which of the following values of $r$, the value of $^n{C_r}$ will be 15

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