In the Olympic women's skating competition, the gold medal goes to first place, silver to second, and bronze to third. If there are 19 skaters, including 2 Americans and 5 Canadians, in how many ways can the medals be awarded to three of the 19 skaters so that an American wins a medal, and a Canadian wins a medal?

ABJeIIy Jun 20, 2024

#1**0 **

We're dealing with a scenario where both American and Canadian skaters need to shine. Here's how we can find the number of winning combinations:

American medalists: There are 2 American skaters, so there are 2 ways to choose one for a medal.

Canadian medalists: There are 5 Canadian skaters, so there are 5 ways to pick one for a medal.

Remaining medal: Once we fix the American and Canadian winners, the third medal can go to any of the remaining 19 skaters minus the 2 Americans and 1 Canadian who already have medals (19 - 2 - 1 = 16 skaters).

So, the total number of favorable outcomes is:

Number of American medalists * Number of Canadian medalists * Number of remaining medalists

2 ways * 5 ways * 16 ways = 160 ways

There you go! 160 different scenarios where both an American and a Canadian skater can snag those prestigious medals.

derpmcfearson Jun 20, 2024