Question
If ${^\text{n+1}}\text{C}_{\text{3}}=2.{^\text{n}}\text{C}_{\text{2}},$ then n:
- 3
- 4
- 5
- 6
Solution:
${^\text{n+1}}\text{C}_{\text{3}}=2\times{^\text{n}}\text{C}_{\text{2}}$
$\Rightarrow \frac{(\text{n}+1)!}{3!(\text{n-2})!}=2\times\frac{\text{n}!}{2!(\text{n}-1)!}$
$\Rightarrow \text{n+1}=6$
$\Rightarrow \text{n}=5$
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If $\frac{(1-3\text{P})}{2},\frac{(1+4\text{P})}{3},\frac{(1+\text{P})}{6}$ are the probabilities of three mutually exclusive and exhaustive events, then the set of all values of p is: