MCQ
If ${^\text{20}}\text{C}_{\text{r}}={^\text{20}}\text{C}_{\text{r+4}}$ is then ${^\text{r}}\text{C}_{\text{3}}$ equal to:
  • A
    $54$
  • $56$
  • C
    $58$
  • D
    none of these.

Answer

Correct option: B.
$56$
$\text{r}+\text{r}+4=20$
$\Rightarrow 2\text{r}+4=20$
$\Rightarrow 2\text{r}=16$
$\Rightarrow \text{r}=8$
Now,
${^\text{r}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$
$\therefore\ {^\text{8}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$
$\therefore\ {^\text{8}}\text{C}_{\text{3}}=\frac{8!}{3!5!}$
$=\frac{8\times7\times6}{3\times2\times1}$
$=56$

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