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

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(C) We know the formula $^nC_r = \frac{n!}{r!(n-r)!}$.First,consider the ratio $\frac{^nC_r}{^nC_{r-1}} = \frac{84}{36} = \frac{7}{3}$.Using the prope

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$3$

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(C) We know the formula $^nC_r = \frac{n!}{r!(n-r)!}$.First,consider the ratio $\frac{^nC_r}{^nC_{r-1}} = \frac{84}{36} = \frac{7}{3}$.Using the property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{n-r+1}{r} = \frac{7}{3} \implies 3n - 3r + 3 = 7r \implies 3n - 10r = -3$ (Equation $1$).Next,consider the ratio $\frac{^nC_{r+1}}{^nC_r} = \frac{126}{84} = \frac{3}{2}$.Using the property $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1}$,we get $\frac{n-r}{r+1} = \frac{3}{2} \implies 2n - 2r = 3r + 3 \implies 2n - 5r = 3$ (Equation $2$).Multiply Equation $2$ by $2$: $4n - 10r = 6$ (Equation $3$).Subtract Equation $1$ from Equation $3$: $(4n - 10r) - (3n - 10r) = 6 - (-3) \implies n = 9$.Substitute $n = 9$ into Equation $2$: $2(9) - 5r = 3 \implies 18 - 5r = 3 \implies 5r = 15 \implies r = 3$.

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

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