MCQ
If $^\text{ⁿ}\text{C}_{15} = \ ^\text{ⁿ}\text{C}_6 $then the value of$\ ^\text{ⁿ}\text{C}_{21}$ is:
  • A
    0
  • B
    1
  • C
    21
  • D
    None of these

Answer

  1. 0

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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