If ^{2n}C_2 : ^nC_2 = 9:2 and ^nC_r = 10,then | vedclass

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(B) Given the ratio $^{2n}C_2 : ^nC_2 = 9:2$,we use the formula $^nC_r = \frac{n!}{r!(n-r)!}$.$\frac{\frac{(2n)!}{2!(2n-2)!}}{\frac{n!}{2!(n-2)!}} = \

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$2$

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(B) Given the ratio $^{2n}C_2 : ^nC_2 = 9:2$,we use the formula $^nC_r = \frac{n!}{r!(n-r)!}$.$\frac{\frac{(2n)!}{2!(2n-2)!}}{\frac{n!}{2!(n-2)!}} = \frac{9}{2}$$\frac{(2n)(2n-1)}{n(n-1)} = \frac{9}{2}$$2(2n)(2n-1) = 9n(n-1)$$4(2n-1) = 9(n-1)$ (since $n 
eq 0$)$8n - 4 = 9n - 9$$n = 5$Now,substitute $n=5$ into $^nC_r = 10$:$^5C_r = 10$$\frac{5!}{r!(5-r)!} = 10$Since $^5C_2 = \frac{5 	imes 4}{2 	imes 1} = 10$,we have $r = 2$.

If $^{2n}C_2 : ^nC_2 = 9:2$ and $^nC_r = 10$,then $r = $

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