If {}^n C_{r-1}=36, {}^n C_r=84, and {}^n C_{ | vedclass

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(C) We know the property of binomial coefficients: $\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{n-r+1}{r}$.Using this, we have:$1) \frac{{}^n C_r}{{}^n C_{r

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$9$

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(C) We know the property of binomial coefficients: $\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{n-r+1}{r}$.Using this, we have:$1) \frac{{}^n C_r}{{}^n C_{r-1}} = \frac{84}{36} = \frac{7}{3}$ $\Rightarrow \frac{n-r+1}{r} = \frac{7}{3}$ $\Rightarrow 3n - 3r + 3 = 7r$ $\Rightarrow 3n - 10r = -3$ (Eq. $1$)$2) \frac{{}^n C_{r+1}}{{}^n C_r} = \frac{126}{84} = \frac{3}{2}$ $\Rightarrow \frac{n-(r+1)+1}{r+1} = \frac{3}{2}$ $\Rightarrow \frac{n-r}{r+1} = \frac{3}{2}$ $\Rightarrow 2n - 2r = 3r + 3$ $\Rightarrow 2n - 5r = 3$ (Eq. $2$)Multiplying Eq. $2$ by $2$, we get $4n - 10r = 6$ (Eq. $3$).Subtracting Eq. $1$ from Eq. $3$: $(4n - 10r) - (3n - 10r) = 6 - (-3) \Rightarrow n = 9$.Substituting $n=9$ into Eq. $2$: $2(9) - 5r = 3$ $\Rightarrow 18 - 5r = 3$ $\Rightarrow 5r = 15$ $\Rightarrow r = 3$.Thus, ${}^n C_8 = {}^9 C_8 = {}^9 C_{9-8} = {}^9 C_1 = 9$.

If ${}^n C_{r-1}=36$, ${}^n C_r=84$, and ${}^n C_{r+1}=126$, then the value of ${}^n C_8$ is

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