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Part A. There are $15$ skaters in the Olympic women's competition, including $4$ Americans. The gold medal goes to first place, silver to second, and bronze to third. In how many ways can the medals be awarded to three of the $15$ skaters? Part B There are $15$ skaters in the Olympic women's competition, including $4$ Americans. The gold medal goes to first place, silver to second, and bronze to third. In how many ways can the medals be awarded to three of the $15$ skaters, if exactly one of the Americans wins a medal?

 

Part B. There are  skaters in the Olympic women's competition, including  Americans. The gold medal goes to first place, silver to second, and bronze to third. In how many ways can the medals be awarded to three of the  skaters, if exactly one of the Americans wins a medal?

 Jun 9, 2020
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Part A: The number of ways is 15*14*13 = 2820.

 

Part B:  There are 3 groups of skaters as follows:

 

4 x 11 x 10 =440 ways. But 1st,  2nd and 3rd prizes can be awarded in 3! =6 ways.

 

Therefore, there are: 440 x 6 =2,640 ways.

 Jun 10, 2020

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