If there are 12 runners then we can choose the 1st runner in 12 ways (1 out of 12), the second in 11 ways (1 out of 11 remaining runners because we have already chosen one) and the third we can choose in 10 ways (1 out of (12-2)=10).

So if there are 12 runner they can finish 1st, 2nd and 3rd in

`12cdot11cdot10=1320 "ways"`.

This problem is the same as choosing 3 objects from group of 12 objects.