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|Online Test Name||Permutations and Combinations|
|Exam Type||Multiple Choice Questions|
|Number of Questions||53|
Answer: Option B
Two men, three women and one child can be selected in ⁴C₂ * ⁶C₃ * ⁵C₁ ways
= (4 * 3)/(2 * 1) * (6 * 5 * 4)/(3 * 2) * 5
Answer: Option C
The number of ways of selecting three men, two women and three children is:
Answer: Option C
Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
Answer: Option D
The total number of ways of forming the group of ten representatives is ²²C₁₀.
Answer: Option E
We know that, the number of straight lines that can be formed by the 11 points in which 6 points are collinear and no other set of three points, except those that can be selected out of these 6 points are collinear.