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7.Binomial Theorem
normal
If $^n{C_{r - 2}} = 36$ , $^n{C_{r - 1}} = 84$ and $^n{C_r} = 126$ , then value of $^n{C_{2r}}$ is
A
$9$
B
$36$
C
$66$
D
$126$
Solution
$^{\rm{n}}{{\rm{C}}_{{\rm{r}} – 2}} = 36,{\,^{\rm{n}}}{{\rm{C}}_{{\rm{r}} – 1}} = 84,{\,^{\rm{n}}}{{\rm{C}}_{\rm{r}}} = 126$
$\frac{{^n{C_{r – 1}}}}{{^n{C_{r – 2}}}} = \frac{{84}}{{36}} \Rightarrow \frac{{{\rm{n}} – {\rm{r}} + 2}}{{{\rm{r}} – 1}} = \frac{7}{3}$
$\Rightarrow 3 \mathrm{n}+13=10 \mathrm{r}$ ….$(1)$
$\frac{{^{\rm{n}}{C_r}}}{{^n{{\rm{C}}_{r – 1}}}} = \frac{{126}}{{84}} \Rightarrow \frac{{{\rm{n}} – {\rm{r}} + 1}}{{\rm{r}}} = \frac{3}{2}$
$\Rightarrow 2 \mathrm{n}+2=5 \mathrm{r}$ ……$(2)$
$\therefore \mathrm{n}=9, \mathrm{r}=4$
$^n{C_{r – 1}} = {\,^9}{{\rm{C}}_8} = 9$
Standard 11
Mathematics
