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

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(B) We use the property $\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}$.For the first two equations:$\frac{^nC_r}{^nC_{r-1}} = \frac{84}{36} = \frac{7}{3}

Vedclass · Accepted Answer

$9$

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(B) We use the property $\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}$.For the first two equations:$\frac{^nC_r}{^nC_{r-1}} = \frac{84}{36} = \frac{7}{3} \implies \frac{n-r+1}{r} = \frac{7}{3} \implies 3n - 3r + 3 = 7r \implies 3n - 10r = -3$  $(1)$For the next two equations:$\frac{^nC_{r+1}}{^nC_r} = \frac{126}{84} = \frac{3}{2} \implies \frac{n-(r+1)+1}{r+1} = \frac{3}{2} \implies \frac{n-r}{r+1} = \frac{3}{2} \implies 2n - 2r = 3r + 3 \implies 2n - 5r = 3$  $(2)$Multiply equation $(2)$ by $2$ to get $4n - 10r = 6$  $(3)$.Subtract equation $(1)$ from $(3)$:$(4n - 10r) - (3n - 10r) = 6 - (-3) \implies n = 9$.

If $^nC_r = 84$,$^nC_{r-1} = 36$,and $^nC_{r+1} = 126$,then $n = ..........$

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