If ^nC_{r-2} = 36,^nC_{r-1} = 84,and ^nC_r = | vedclass

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(A) Given: $^nC_{r-2} = 36$,$^nC_{r-1} = 84$,and $^nC_r = 126$.Using the property $\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}$:$\frac{^nC_{r-1}}{^nC_{r

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$9$

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(A) Given: $^nC_{r-2} = 36$,$^nC_{r-1} = 84$,and $^nC_r = 126$.Using the property $\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}$:$\frac{^nC_{r-1}}{^nC_{r-2}} = \frac{84}{36} = \frac{7}{3}$ $\Rightarrow \frac{n-(r-1)+1}{r-1} = \frac{7}{3}$ $\Rightarrow \frac{n-r+2}{r-1} = \frac{7}{3}$$3n - 3r + 6 = 7r - 7 \Rightarrow 3n - 10r = -13$  ....$(1)$$\frac{^nC_r}{^nC_{r-1}} = \frac{126}{84} = \frac{3}{2} \Rightarrow \frac{n-r+1}{r} = \frac{3}{2}$$2n - 2r + 2 = 3r \Rightarrow 2n - 5r = -2$  ....$(2)$Multiplying equation $(2)$ by $2$: $4n - 10r = -4$  ....$(3)$Subtracting $(1)$ from $(3)$: $(4n - 10r) - (3n - 10r) = -4 - (-13) \Rightarrow n = 9$Substituting $n=9$ in $(2)$: $2(9) - 5r = -2$ $\Rightarrow 18 - 5r = -2$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$We need to find $^nC_{2r} = ^9C_{2(4)} = ^9C_8 = ^9C_1 = 9$.

If $^nC_{r-2} = 36$,$^nC_{r-1} = 84$,and $^nC_r = 126$,then the value of $^nC_{2r}$ is

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