If binom{n}{r-1} = 36,binom{n}{r} = 84,and bi | vedclass

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(C) Given $\binom{n}{r-1} = 36$,$\binom{n}{r} = 84$,and $\binom{n}{r+1} = 126$.Taking the ratio of the first two terms:$\frac{\binom{n}{r}}{\binom{n}{

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(C) Given $\binom{n}{r-1} = 36$,$\binom{n}{r} = 84$,and $\binom{n}{r+1} = 126$.Taking the ratio of the first two terms:$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{84}{36} = \frac{7}{3}$$\frac{n!}{r!(n-r)!} 	imes \frac{(r-1)!(n-r+1)!}{n!} = \frac{7}{3}$$\frac{n-r+1}{r} = \frac{7}{3} \implies 3n - 3r + 3 = 7r \implies 3n - 10r = -3$  (Equation $1$)Taking the ratio of the next two terms:$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{126}{84} = \frac{3}{2}$$\frac{n!}{(r+1)!(n-r-1)!} 	imes \frac{r!(n-r)!}{n!} = \frac{3}{2}$$\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)$$n = 9$Substitute $n=9$ into Equation $2$:$2(9) - 5r = 3 \implies 18 - 5r = 3 \implies 5r = 15 \implies r = 3$.

If $\binom{n}{r-1} = 36$,$\binom{n}{r} = 84$,and $\binom{n}{r+1} = 126$,then $r = \dots$

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