1. **Problem statement:** Three runners compete in a race. We want to find the number of ways they can finish the race under two conditions: a) No tied places. b) Tied places allowed. 2. **Case a: No tied places** When no ties are allowed, each runner must have a unique finishing position (1st, 2nd, 3rd). The number of ways to arrange $n$ distinct items is given by the factorial formula: $$n! = n \times (n-1) \times (n-2) \times \cdots \times 1$$ For 3 runners, the number of ways is: $$3! = 3 \times 2 \times 1 = 6$$ 3. **Case b: Tied places allowed** When ties are allowed, the problem becomes counting the number of weak orders or total preorders on 3 elements. The number of ways to rank $n$ items allowing ties is given by the $n$th ordered Bell number (also called Fubini number). For $n=3$, the ordered Bell number is 13. This counts all possible rankings including all tie combinations. 4. **Summary:** - No ties: 6 ways - Ties allowed: 13 ways Thus, the answers are: $$\boxed{6}$$ ways with no ties $$\boxed{13}$$ ways with ties allowed