If ^n{Cᵣ} = 84,{^n}{C_{r - 1}} = 36 and ^n{C_{r + 1}} = 126 , then n equals

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6.Permutation and Combination

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If $^n{C_r} = 84,{\;^n}{C_{r - 1}} = 36$ and $^n{C_{r + 1}} = 126$, then $n$ equals

A

$8$

B

$9$

C

$10$

D

$5$

Solution

(b) $\frac{{n – r + 1}}{r} = \frac{{84}}{{36}} = \frac{7}{3}$ and $\frac{{n – r}}{{r + 1}} = \frac{{126}}{{84}} = \frac{3}{2}$

$\therefore $$\frac{7}{3}r – 1 = n – r = \frac{3}{2}(r + 1)$

or $14r – 6 = 9r + 9$ or $r = 3$. So, $n = 9$.

Standard 11

Mathematics

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