MCQ
If ${^\text{20}}\text{C}_{\text{r}}={^\text{20}}\text{C}_{\text{r-10}}$ is then ${^\text{18}}\text{C}_{\text{r}}$ equal to:
- A4896
- B816
- C1632
- DNone of these.
Solution:
$\text{r}+\text{r}-10=20$
$\Rightarrow 2\text{r}-10=20$
$\Rightarrow 2\text{r}=30$
$\Rightarrow \text{r}=15$
Now,
${^\text{18}}\text{C}_{\text{r}}={^\text{18}}\text{C}_{\text{15}}$
$\therefore\ {^\text{18}}\text{C}_{\text{15}}={^\text{18}}\text{C}_{\text{3}}$
$\therefore\ {^\text{18}}\text{C}_{\text{3}}=\frac{18}{3}\times\frac{17}{2}\times16$
$=816$
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