MCQ
If ${^\text{m}}\text{C}_{\text{1}}={^\text{n}}\text{C}_{\text{2}},$ is then:
- A2m = n
- B2m = n(n + 1)
- C2m = n(n - 1)
- D2n = m(m - 1)
Solution:
${^\text{m}}\text{C}_{\text{1}}={^\text{n}}\text{C}_{\text{2}}$
$\Rightarrow \frac{\text{m!}}{1!(\text{m}-1)!}=\frac{\text{n!}}{2!(\text{n}-2)!}$
$\Rightarrow \frac{\text{m}(\text{m}-1)!}{(\text{m}-1)!}=\frac{\text{n}(\text{n}-1)(\text{n}-2)!}{2!(\text{n}-2)!}$
$\Rightarrow2\text{m}=\text{n}(\text{n}-1)$
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