Suppose 31 cars start at a car race. In how many ways can the top 3 cars finish the race? The number of different top three finishes possible for this race of 31 cars is (Use integers for any number in the expression.)

Step 1 ：Suppose 31 cars start at a car race. We want to find out in how many ways the top 3 cars can finish the race.

Step 2 ：This is a permutation problem. We have 31 cars and we want to know how many ways we can arrange 3 of them. The formula for permutations is: \(P(n, r) = \frac{n!}{(n-r)!}\) where n is the total number of items, r is the number of items to choose, and '!' denotes factorial.

Step 3 ：In this case, n = 31 (the total number of cars) and r = 3 (the number of cars to finish the race). So we need to calculate \(P(31, 3)\).

Step 4 ：By substituting the values into the formula, we get \(P(31, 3) = \frac{31!}{(31-3)!}\)

Step 5 ：After calculating the above expression, we find that the number of different top three finishes possible for this race of 31 cars is 26970.

Step 6 ：Final Answer: \(\boxed{26970}\)