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I'm not sure how to solve the questions in my math class because I can't grasp the logic behind the story.

For example given the following questions:

There are 8 men and 6 women. 4 Men and 2 women will be selected to form a team, in how many ways can the teams be arranged?

I would answer the above question with the following solution: $\binom{8}{4} * \binom{6}{2} = \frac{8!}{(4! \centerdot 4!)} * \frac{6!}{(4! \centerdot 2!)} = 1050$. Which I think is correct.

But with problems like:

In a competition with 10 competitors, in how many ways can 3 medals be shared among them?

I would answer this question the same way I answered the above question with: $\binom{10}{4}$. To my surprise this seems to be incorrect, it's probably because I can't get a hold on the logic involved.

It would be great if the logic behind the problems can be explained in some detail.

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    $\begingroup$ If the three medals are identical (and if you replace 10-choose-4 by 10-choose-3), your answer is correct. But if one distinguishes between the medals (gold, silver and bronze), there are 6 times more different configurations. $\endgroup$
    – Did
    Commented Apr 12, 2012 at 9:06

1 Answer 1

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The key difference between two problems is that medals are considered to be different (e.g. golden medal, silver medal etc), while positions in team are the same (that is, it doesn't matter whether a specific women will be on the first position or on the second).

Thus, there are $10*9*8 = \frac{10!}{7!}$ ways to distribute three medals between 10 competitors (e.g. golden medal goes to #5, silver medal goes to #7, bronze medal goes to #2), but only $\frac{10*9*8}{3!} = \binom{10}{3}$ ways to choose what three competitors will receive the medals (e.g. #2, #5 and #7 are getting some medals / positions / etc).

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