# 8 Athletes run a race problem

8 Athletes run a race and no two athletes finish exactly together. What is the number of different possible results for the first, second and third positions?

I said that the answer is simply 8 choose 3, which gives an answer of 56. When I checked the memo, it said the answer was 336. How is this?

• The athletes ar distingushabel. So Joe, Sam, Jim is not the same as San, Joe, Jim. Am I right?
– zoli
Mar 13 '15 at 13:26
• @zoli: If you cannot distinguish the athletes, what would be the point of having them compete in a race? Mar 13 '15 at 13:34
• Mark: Right. Right.
– zoli
Mar 13 '15 at 13:39

8 choose 3 is the number of ways of choosing 3 different runners from 8 but you haven't accounted for the fact that once you've picked three, they can come in a number of different orders, specifically the number of permuations of 3 elements (i.e. 3!=6). So the final answer is $$\binom{8}{3}\times 3! = 56\times 6 = 336$$

• Oh ok, thank you for the explanation. I thought that when you combined the athletes you would automatically get the different permutations of 1st, 2nd and 3rd position. Mar 14 '15 at 6:51

The order in which the runners finish matters, so this is a permutation rather than a combination.

Until the first runner crosses the finish line, there are eight runners who could finish first. Once the first runner crosses the finish line, there are seven runners who could finish second. Once the first two runners have crossed the finish line, there are six runners who could finish third, giving $$8 \cdot 7 \cdot 6 = 336$$ ways for the runners to finish first, second, and third in the race given that there are no ties. Using permutations, the number of ways the eight runners could finish first, second, and third given that there are no ties is $$P(8, 3) = \frac{8!}{(8 - 3)!} = \frac{8!}{5!} = \frac{8 \cdot 7 \cdot 6 \cdot 5!}{5!} = 8 \cdot 7 \cdot 6 = 336$$

Imagine if John, Jack and Julie were in the "top-3", there is 6 possible combination.Because you do not know what was the rank. Anyone can be at

So $8\choose3$ ways of selecting 3 people out of a group of 8, multiplied by no of ways of arranging those 3 in the ranks $3!$.