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

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(B) We are given the following relations for combinations:$1$) $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} = \frac{84}{36} = \frac{7}{3}$$2$) $\frac{^n

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$9$

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(B) We are given the following relations for combinations:$1$) $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} = \frac{84}{36} = \frac{7}{3}$$2$) $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1} = \frac{126}{84} = \frac{3}{2}$From $(1)$,$3(n-r+1) = 7r \implies 3n - 3r + 3 = 7r \implies 3n - 10r = -3$From $(2)$,$2(n-r) = 3(r+1) \implies 2n - 2r = 3r + 3 \implies 2n - 5r = 3$Multiply the second equation by $2$: $4n - 10r = 6$Subtract the first equation from this: $(4n - 10r) - (3n - 10r) = 6 - (-3) \implies n = 9$Substituting $n=9$ into $2n - 5r = 3$,we get $18 - 5r = 3 \implies 5r = 15 \implies r = 3$.

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

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