MCQ
If ${^\text{15}}\text{C}_{3\text{r}}={^\text{15}}\text{C}_{\text{r+3}},$ is then equal to:
- A5
- B4
- C3
- D2
Solution:
$3\text{r}+\text{r}+3=15$
$\Rightarrow 4\text{r}+3=15$
$\Rightarrow 4\text{r}=12$
$\Rightarrow \text{r}=3$
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The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1: