+1678 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?
+81 This person isn't very good at combinatorics, but I guess I will give this a try? 
Let's call the American skaters \(a\), the Canadian skaters \(c\), and the other skaters \(o\).
(Quick note, here are the numbers for \(a,c,\) and \(o \).)
\(a=2\)
\(c=5\)
\(o=12\)
Than, in the top 3 skaters, we want a standing that looks something like this: \((a_1,o_6,c_2)\)
(for this problem, I'm assuming that the Americans or Canadians are different, let me know otherwise...)
We can think just think of one case, since we will be multiplying the probablities, and multiplication is commutative.
(BTW, let me know if there is AT LEAST ONE in the problem, or just one American/Canadian)
I'm just gonna take the case \((a,c,o)\). For the American, there are 2 choices: \(a_1 \) and \(a_2\). For the Canadians, likewise, there are 5 choices. And for the other 12 people, there are 12 choices. We can simply multiply these 3 numbers together and get \(2*5*12 = 120\). We can multiply this number by 6, since there are \(3! \) ways to arrange the standings, so we get \(120*6 = \boxed{720}\)
+81 This person isn't very good at combinatorics, but I guess I will give this a try?
Let's call the American skaters \(a\), the Canadian skaters \(c\), and the other skaters \(o\).
(Quick note, here are the numbers for \(a,c,\) and \(o \).)
\(a=2\)
\(c=5\)
\(o=12\)
Than, in the top 3 skaters, we want a standing that looks something like this: \((a_1,o_6,c_2)\)
(for this problem, I'm assuming that the Americans or Canadians are different, let me know otherwise...)
We can think just think of one case, since we will be multiplying the probablities, and multiplication is commutative.
(BTW, let me know if there is AT LEAST ONE in the problem, or just one American/Canadian)
I'm just gonna take the case \((a,c,o)\). For the American, there are 2 choices: \(a_1 \) and \(a_2\). For the Canadians, likewise, there are 5 choices. And for the other 12 people, there are 12 choices. We can simply multiply these 3 numbers together and get \(2*5*12 = 120\). We can multiply this number by 6, since there are \(3! \) ways to arrange the standings, so we get \(120*6 = \boxed{720}\)
