$^{n} C_{r - 1} = 330,^{n} C_{r} = 462,^{n} C_{r + 1} = 462 \Rightarrow r =$

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Solution

The correct option is D 5
$^{n} C_{r - 1} = 330$ , $^{n} C_{r} =^{n} C_{r + 1} = 462$

$\Rightarrow \frac{n!}{(r!) (n - r)!} = \frac{n!}{(r + 1)! (n - (r + 1))!}$

$\Rightarrow (r + 1)! (n - (r + 1))! = (r!) (n - r)!$

$\Rightarrow r + 1 = n - r$

$n = 2 r + 1$

$\frac{n - r + 1}{r} = \frac{462}{330}$

$= 330 n - 330 r + 330 = 462 r$

$= 330 n + 330 = 792 r$

$= (n + 1) 330 = 792 r$

$= (2 r + 2) 330 = 792 r$

$= 660 r + 660 = 792 r$

$= 660 = 132 r$

$\Rightarrow r = \frac{660}{132}$

$\to r = 5$

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Similar questions

$^{n} C_{r - 1} = 330,^{n} C_{r} = 462,^{n} C_{r + 1} = 462 \Rightarrow r =$

\sum_{r = 0}^{n - 1} \frac{n_{C_{r}}}{n_{C_{r}} + n_{C r + 1}}

^{n} C_{r - 1} = 36,^{n} C_{r} = 84,^{n} C_{r + 1} = 126

(n, r)

=

^{n} C_{r - 1} = 36,^{n} C_{r} = 84

^{n} C_{r + 1} = 126,

r

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