MCQ
If $^\text{ⁿ}\text{C}_{15} = \ ^\text{ⁿ}\text{C}_6 $then the value of$\ ^\text{ⁿ}\text{C}_{21}$ is:
- A0
- B1
- C21
- DNone of these
Solution:
We know that
if $\ ^\text{ⁿ}\text{Cr}_1 =\ ^ⁿ\text{Cr}_2$
$\Rightarrow\text{n}= \text{r}^1 + \text{r}^2$
Given, $^\text{ⁿ}\text{C}_{15} = \ ^\text{ⁿ}\text{C}_6 $
$\Rightarrow\text{n} = 15 + 6$
$\Rightarrow\text{n} = 21$
Now, $ ^{21}\text{C}_{21} = 1$
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