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In a race, at least 3 and at most 5 runners will
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21 Aug 2017, 12:33
21 Aug 2017, 12:33
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Question Stats:
54% (01:18) correct
45% (01:29) wrong based on 130 sessions
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In a race, at least 3 and at most 5 runners will vie for gold, silver, and bronze medals. Which of the following could represent the total number of unique ways to distribute the three medals among the participants?
Indicate all such numbers.
❑ 3
❑ 6
❑ 12
❑ 24
❑ 30
❑ 60
Re: In a race, at least 3 and at most 5 runners will
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24 Sep 2017, 07:39
24 Sep 2017, 07:39
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The crucial information to which paying attention is unique ways. Let's analyze the three occasions:
- if the runners are 3, there are 3! ways to allocate 3 medals among 3 people: 3*2*1 = 6 - Answer B
- if the runner are 4, we have to consider 4C3 to get the number of combinations without paying attention to the order, but since we are asked to count the unique ways, we have to multiply it by 3! in order to consider each variant as unique: 4C3*3!=4*3*2*1 = 24 - Answer D
- if runners are 5, the argument is above, we have to take 5C3*3! = 5!/(2!*3!)*3! = 10*3*2*1 = 60 - Answer F
- if the runners are 3, there are 3! ways to allocate 3 medals among 3 people: 3*2*1 = 6 - Answer B
- if the runner are 4, we have to consider 4C3 to get the number of combinations without paying attention to the order, but since we are asked to count the unique ways, we have to multiply it by 3! in order to consider each variant as unique: 4C3*3!=4*3*2*1 = 24 - Answer D
- if runners are 5, the argument is above, we have to take 5C3*3! = 5!/(2!*3!)*3! = 10*3*2*1 = 60 - Answer F
Re: In a race, at least 3 and at most 5 runners will
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21 Feb 2018, 13:23
21 Feb 2018, 13:23
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Each metal is different than the others.
Therefore, we can use 3P3,4P3 and 5P3 for the computation.
3p3=6 4p3=24 5p3=60
Therefore, we can use 3P3,4P3 and 5P3 for the computation.
3p3=6 4p3=24 5p3=60
