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

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(C) We are given the following equations:$(1)$ $\frac{^nC_{r-1}}{^nC_r} = \frac{36}{84} = \frac{3}{7}$Using the formula $\frac{^nC_{r-1}}{^nC_r} = \fr

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(C) We are given the following equations:$(1)$ $\frac{^nC_{r-1}}{^nC_r} = \frac{36}{84} = \frac{3}{7}$Using the formula $\frac{^nC_{r-1}}{^nC_r} = \frac{r}{n-r+1}$,we get $\frac{r}{n-r+1} = \frac{3}{7} \implies 7r = 3n - 3r + 3 \implies 3n - 10r = -3$.$(2)$ $\frac{^nC_r}{^nC_{r+1}} = \frac{84}{126} = \frac{2}{3}$Using the formula $\frac{^nC_r}{^nC_{r+1}} = \frac{r+1}{n-r}$,we get $\frac{r+1}{n-r} = \frac{2}{3} \implies 3r + 3 = 2n - 2r \implies 2n - 5r = 3$.Multiplying the second equation by $2$,we get $4n - 10r = 6$.Subtracting the first equation $(3n - 10r = -3)$ from this,we get $n = 9$.Substituting $n = 9$ into $2n - 5r = 3$,we get $18 - 5r = 3 \implies 5r = 15 \implies r = 3$.

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

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