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  4. If nCr-1=36, nCr=84 and nCr+1=126 then the value of nC8

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Q. If $^{n}C_{r-1}=36, ^{n}C_{r}=84$ and $^{n}C_{r+1}=126 $ then the value of $^{n}C_{8}$

WBJEEWBJEE 2016Permutations and Combinations

Solution:

$\frac{n}{r-1\, n-r+1}=36 .....\left(1\right)$
$\frac{n}{r\, n-r}=84 ...........\left(2\right)$
$\frac{n}{r+1\, n-r-1}=126 ......\left(3\right)$
$\left(1\right) \div\left(2\right)$ gives $\frac{r}{n-r+1}=\frac{36}{84} \Rightarrow 84r=36n-36r+36$ or $120r=36n+36 ......\left(4\right)$
$\left(2\right)\div\left(3\right)$ gives $\frac{r+1}{n-r}=\frac{84}{126} \Rightarrow 126r+126=84n-84r$ or$210r=84n-126 ...........\left(5\right)$
Solving $\left(4\right)$ and $\left(5\right) n = 9, r = 3$
So $^{n}C_{8}=^{9}C_{8}=9$