There are 19 figure skaters in the Olympic women's competition, including 3 Americans. The gold medal goes to first place, silver to second,

Hi Geno, I was looking at what you did.

If you choose the non US prize winners like you did that is 16*15 then you are saying that there are 240 permutaions of those 2 places.

then 1 us person must be chosen.  this can be done in 3 ways

now that US person can be 1st 2nd or 3rd     So 3 ways

240*3*3 = 2160   

You should not multiply this by 2 as all the possible permutations of the non US prize winners in relation to each other have already been accounted for.

There are three different choices for the American.

There are 16 different choices for the first non-American and 15 different choices for the other non-American.

So, there are 3 x 16 x 15 different groups.

But, order is important -- who gets the gold, who gets the silver, and who gets the bronze -- there are 6 different ways to arrange the 3 winners (3! = 6).

Thus, there are 6 x 3 x 16 x 15 different results.

I know my answer is incorrect and geno's is correct.....but can someone spot my error??

I chose to see the problem this way.....

In each set of three people, we want to choose any 1 of 3 Americans and then choose any 2 of 16 of the other contestants

So the total possible sets - disregarding order, is  C(3,1)*C(16,2) = 3 * 120  = 360

And each of these sets can be ordered in 3!  = 6 ways

So......6 x 360 = 2160  which is 1/2 of geno's result.......????....where did i go wrong???

CPhill -- I think the problem arises by choosing two groups, the groups of Americans and the groups of non-Americans.

By doing the problem this way, there are two arrangements, either choosing the Americans first and then the non-Americans, or vice-versa. Then, there are the six arrangements of the individuals. This would double your answer.

Or, maybe not ... ?

Why do you think that you are wrong Chris ? I think that yours is the correct solution.

Geno's solution is wrong because it orders the non-Americans twice. The 16×15 is the number of permutations of 2 from 16 and so is already taking order into account. If simply two non-Americans were required, then C (16, 2)=16×15/2 should be used.

Here's an alternative way of arriving at the result

The number of ways in which America wins a single medal, gold = 3×16×15 = 720.

The number of ways in which America wins a single medal, silver = 16×3×15 = 720.

The number of ways in which America wins a single medal, bronze = 16×15×3 = 720.

The total is 720 + 720 + 720 = 2160.

OK...thanks, Melody....you know....I'm NEVER too sure about my answers to any probability question....!!!!

Why are you talking like I wrote that last annonymous comment Chris?

I did not write that.    

However,

I just looked at the question and I agree with you and anon.  That is how I would have done it.

I am not saying it is necessarily correct but it does make sense to me.     

Sorry....I misread that one.......maybe I'm still dizzy....LOL!!!

I do not think that anyone on this forum is really strong when it comes to probability.

I do however think that your knowledge has been improving very rapidly in this subject area.  Now all you really need is a bit more confidence.  

....Or a bit more wine......

Yes that might help too - might help me too LOL  :))