Question

There is a 2019 Math Competition that had 47 participants. The contest provided $500 for 1st place, $200 for 2nd place, and $100 for 3rd place. How many unique groups of winners are possible?

Answer 1

For 1st 2nd and 3rd place we have to choose 3 participants from the 47 participants . Here ordering matters. That means {1,2,3} and {2,1,3} are not same.

So for the 1st position there is total 47 participants. That means 47 different ways . After choosing 1st one ,for the 2nd place there is remaining 46 participants. Because the 1st place holder can't take the 2nd place, so for second place we have 46 different ways . And finally for 3rd place we have remaining 45 participants. That means 45 different ways.

So total number of different possible groups are

= 47×46×45 = 97290

Or you can simply find this by = 47P3 = 97290