MCQ
If ${^\text{n+1}}\text{C}_{\text{3}}=2.{^\text{n}}\text{C}_{\text{2}},$ then n:
  • A
    3
  • B
    4
  • 5
  • D
    6

Answer

Correct option: C.
5
${^\text{n+1}}\text{C}_{\text{3}}=2\times{^\text{n}}\text{C}_{\text{2}}$
$\Rightarrow \frac{(\text{n}+1)!}{3!(\text{n-2})!}=2\times\frac{\text{n}!}{2!(\text{n}-1)!}$
$\Rightarrow \text{n+1}=6$
$\Rightarrow \text{n}=5$

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