MCQ
If ${^\text{n+1}}\text{C}_{\text{3}}=2.{^\text{n}}\text{C}_{\text{2}},$ then n:
- A3
- B4
- C5
- D6
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 line joining (1, 2) and (5, 7) is parallel to line joining (3, 4) and (11, x):
The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y - 11 = 0 are: