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

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

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(b) $\frac{{n - r + 1}}{r} = \frac{{84}}{{36}} = \frac{7}{3}$ and $\frac{{n - r}}{{r + 1}} = \frac{{126}}{{84}} = \frac{3}{2}$

$	herefore $$\frac{7}{3}r - 1 = n - r = \frac{3}{2}(r + 1)$

or $14r - 6 = 9r + 9$ or $r = 3$. So, $n = 9$.

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

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