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?
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?
