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If $^nC_{r-1}=36$, $^nC_r=84$ an $^nC_{r+1}=126$ then $r$ is equal to

  1. $1$
  2. $2$
  3. $3$
  4. none of these
in Combinatory by Veteran (424k points)
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We can go with  options also.

Here $n=9,r=3$

1 Answer

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Answer: $\mathbf C$

Explanation:

$\binom{n}{r} = 84 \implies \frac{n!}{(n-r)!r!} = 84$

$\binom{n}{r-1} = 84 \implies \frac{n!}{(n-r+1)!(r-1)!} = 84$

$\binom{n}{r+1} = 84 \implies \frac{n!}{(n-r-1)!(r+1)!} = 126$

$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{84}{36} = \frac{7}{3} \implies \frac{n!}{(n-r)!r!} .\frac{(n-r+1)!(r-1)!}{n!} = \frac{n-r+1}{r} = \frac{7}{3}$

On simplifying this, we get:

$3n + 3 = 10r \cdots(1)$

Similarly,

$\frac{\binom{n}{(r+1)}}{\binom{n}{r}}=\frac{126}{84} = \frac{3}{2} \implies \frac{n-r}{r+1} = \frac{3}{2}$

Again, on simplifying this, we get:

$2n-3 = 5r \cdots(2)$

On solving equation $(1)$ and $(2)$, we get:

$r= 3$

$\therefore \mathbf C$ is the correct option.

by Boss (12.9k points)
edited by

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