Problem Solving

The explanation in the book seems way too complicated so hopefully someone can explain this to me a little easier.

Fifteen runners from four different countries are competing in a tournament. Each country holds a qualifying heat to determine who its fastest runner is. These four runners then run a final race for first, second, and third place. If no country has more than one more runner than any other country, how many arrangements of prize-winners are there?

24
384
455
1248
2730


This question was addressed in a post on a diff website and the explanation is:

My approach:
1. We have runners by countries: 4 4 4 3
2. We choose 3 countries for all 3 "wining" places. 4 possibility: 4 4 4 and 3 times 4 4 3
3. Now we can choose any person from each country and take into account different positions (3!)

So, we get: (4*4*4)*3! + 3*(4*4*3)*3! = 3!*4*4*(4+3*3) = 1248 (D)

Marco83, a good question!
+1


But I just do not understand the logic behind what he is talking about....can someone explain this problem's solution a little clearer for me?

ps - combination/permutation/probability are far and away my hardest types of quant questions and I have logged manyyyy hours and have referenced a number of different sites/threads but do not seem to be getting much better at the 700-800 level questions....are these generally considered the hardest questions on the gmat?

Yeah probability questions are considered a bit on difficult side for GMAT.You can however solve these questions if you approach them logically.Every sentence in this question is giving an information which we can't ignore.

no country has more than one more runner than any other country
So fifteen runners can be distributed into 4,4,4 and 3 runners.If you consider any other distribution,you will violate above condition.
Now,these runners are participating in their national qualifying heat and only fastest of them are eligible to run in the international race.So,from four countries,these four runners then run a final race for first, second, and third place.
Cases possible:
Case 1.Runner from the nation having three runners initially come last and first, second, and third place went to runners who already ran in group of four.
Now,fastest runner from each nation can be selected in 4 ways each.
Total ways = 4*4*4
Now these runner's winning position can be arranged in 3! ways(viz. 123,132,213,231,312,321)
Total arrangement possible=3! 4*4*4 =384
Case 2 Runner from the nation having three runners initially didn't come last .So,one nation having four runners initially came last.
Now,we can select two nations of four runners from three nations of four in 3C2 ways=3 ways
Now,fastest runner from each nation can be selected in 4,4,3 ways each.
Total ways=4*4*3
Now these runner's winning position can be arranged in 3! ways
Total arrangements possible=3*4*4*3*3!
= 864
Adding,we get 384+864=1248