kvcpk wrote:GMATGuruNY wrote:Be careful here. The final round is immaterial. There are 8 contestants. Couldn't any one of these 8 be among the final 3 medal winners? All we need to determine is the the number of combinations of 3 that can be made from the 8 contestants:
(8*7*6)/(1*2*3)=56.
Thanks for the correction Mitch!!
Happy to help!
Here's what the arithmetic you performed above represents:
8c6 = the number of combinations of 6 that can be made from 8 choices = 28
6c3 = the number of combinations of 3 that can be made from 6 choices = 20
28 * 20 = 560 determines the number of ways we can COMBINE the combinations of 6 with the combinations of 3. It's as if we had 28 shirts and 20 ties and we we were determining how many ways we could combine them to make an outfit.
One more piece of advice. Given that we have only 8 contestants total, 560 is a huge number of combinations. Before you choose an answer choice, always ask yourself whether it makes sense given the constraints of the problem.
Last edited by GMATGuruNY on Thu Jul 01, 2010 8:16 am, edited 1 time in total.
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