be a contest manager & calculate combinations

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In a group of 8 semifinalists, all but 2 will advance to the final round. If in the final round only the top 3 will be awarded the medals, then how many groups of medal winners are possible?

Since only 6 will advance to the final round i found no of groups possible for 3 out of 6 contestants i.e 6!/3!(6-3)!=20

but this answer is wrong can someone suggest what is wrong??
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by chris500 » Tue Sep 09, 2008 5:52 am
I was sort of confused by this, but I believe this is the answer and how to get it.

The numbers you want to work with are 8 and 3; 6 is thrown in to mess with you. You have 8 total candidates for the 3 positions available at the end, it doesn't matter how many advance to the semi finals, it will be 6 of the 8 in the pool.

So the formula to solve is

N!
_____
K! (N-K)!

K= the number of final positions to choose
N= the number of people to choose from


So the formula follows as

8!
______
3!(8-3)!

continued

8x7x6x5x4x3x2x1
______
3x2x1(5!)

continued

40320
_______
6(120)

continued

40320
_______
720

= 56