Seven cash prizes of distinct amounts are to be awarded to 3 runners who finish in the first, second and third positions, such that each runner gets at least one prize. In how many ways can it be done if the smallest cash prize received by any runner is greater than the highest cash prize received by any runner who finished after him?

This problem can be solved using the principles of permutations and combinations. Here are the steps:

1. First, we need to distribute the 7 cash prizes among the 3 runners such that each runner gets at least one prize. This can be done in 6 ways: (1,1,5), (1,2,4), (1,3,3), (2,2,3), (2,1,4) and (3,1,3) where the numbers in the brackets represent the number of prizes each runner gets.

2. For each of these distributions, we need to arrange the prizes in descending order such that the smallest cash prize received by any runner is greater than the highest cash prize received by any runner who finished after him.

   For example, if the distribution is (1,1,5), the first runner can get any of the 7 prizes, the second runner can get any of the remaining 6 prizes and the third runner gets the remaining 5 prizes. So, the number of ways is 765 = 210.

3. We need to repeat step 2 for each of the 6 distributions and add up the results to get the total number of ways.

   For (1,2,4), the number of ways is 76543 = 2520.

   For (1,3,3), the number of ways is 76543*2 = 5040.

   For (2,2,3), the number of ways is 76543*2 = 5040.

   For (2,1,4), the number of ways is 76543*2 = 5040.

   For (3,1,3), the number of ways is 76543*2 = 5040.

4. Adding up all these results, we get 210 + 2520 + 5040 + 5040 + 5040 + 5040 = 22890.

So, there are 22890 ways to distribute the 7 cash prizes among the 3 runners under the given conditions.