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by diebeatsthegmat » Thu Dec 22, 2011 10:05 am
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?

A) 24
B) 384
C) 455
D) 1248
E) 2730
i find the final answer is 4608 and its not in the among of answer choices. this is my method:
Choose 4 countries: 4 4 4 3: = 4
suppose 4 countries are a b c d
choose the fastest in a country: 4 and the same with b and c is 4 4 and d is 3
choose fastest, second, third a,b,c,d: 4*3*2
total is 4*4*4*4*3*4*6=4608
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by user123321 » Thu Dec 22, 2011 10:46 am
diebeatsthegmat wrote: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?

A) 24
B) 384
C) 455
D) 1248
E) 2730
i find the final answer is 4608 and its not in the among of answer choices. this is my method:
Choose 4 countries: 4 4 4 3: = 4
suppose 4 countries are a b c d
choose the fastest in a country: 4 and the same with b and c is 4 4 and d is 3
choose fastest, second, third a,b,c,d: 4*3*2
total is 4*4*4*4*3*4*6=4608
[spoiler]IMO D?[/spoiler]

We know that these 4 teams(say A,B,C,D) should have 4,4,4,3 members to follow the condition in bold.
lets assume persons from each of A,B,C teams went into final and won
then arrangements for that is 4*4*4*3!
similarly if persons from each of B,C,D teams went into final and won
then arrangements for that is 4*4*3*3!
//ly for C,D,A it is 4*3*4*3!
//ly for D,A,B it is 3*4*4*3!

There are no other ways possible for this so add it up
3!(64+48+48+48) = 6*208 = 1248 ways

The mistake you did is...
say if A1,B1,C1,D1 are winners of prelims then A1,B1,C1 is one arrangement for final winners.
and if A1,B1,C1,D2 are winners of prelims then same A1,B1,C1 can be one arrangement for final winners.

In a way your answer has duplicate arrangements.

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by pemdas » Thu Dec 22, 2011 11:55 am
i believe, the issue is rooted deeper, as diebeatsthegmat tries to understand how it's possible for the 192 ways of arranging runners through four countries (under given conditions) not to be permuted in the right order, like 4*4*4*3 multiplied by 4P3, correct?

>> for essential understanding of 4*4*4*3 simply substitute with dices or codes, the same branching is used here <<

now, please note that we could be in much more predicament if the number of countries were not 4 but 5 or 6!

Compare plz, permutation of 4 by 3 (ordered, as expected by diebeatsthegmat)

x....x x 0
_... _..._..._
0... x...x...x

now it must be evident that three ordered arrangements (1-2-3) put in four slots can be with the left/right sides beginning/ending. This was simple example with four slots (countries); if the number of countries were 5 or 6, we would experience the bigger trouble. Therefore plz do as user suggests

permute each and every branch on its own
3!(4*4*4+4*4*3+4*4*3+4*4*3)
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by GMATGuruNY » Thu Dec 22, 2011 1:18 pm
diebeatsthegmat wrote: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?

A) 24
B) 384
C) 455
D) 1248
E) 2730
The total number of runners = 15.
The total number of countries = 4.
Since no country has more than ONE MORE RUNNER than any other country:
3 countries have 4 runners each (accounting for 12 of the runners).
1 country has 3 runners (bringing the total to 15).
Let A, B, C = the countries with 4 runners each.
Let X = the country with 3 runners.

Case 1: All 3 winners are from A, B, C.
Number of options for first place = 12. (Any of the 12 runners from A, B, C.)
Number of options for 2nd place = 8. (Any of the 8 runners from the 2 remaining countries that did not win 1st place.)
Number of options for 3rd place = 4. (Any of the 4 runners from the 1 remaining country).
To combine these options, we multiply:
12*8*4 = 384.

Case 2: 1 winner from X, 2 winners from A, B, C.
To begin, let's put X in 1st place.
Number of options for 1st place = 3. (Any of the 3 runners from X.)
Number of options for 2nd place = 12. (Any of the 12 runners from A, B, C.)
Number of options for 3nd place = 8. (Any of the 8 runners from the 2 remaining countries.)
To combine these options, we multiply:
3*12*8 = 288.
Since X could be in 1st, 2nd, or 3rd place, we multiply by 3:
3*288 = 864.

Total options = 384 + 864 = 1248.

The correct answer is D.
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by shekhar.kataria » Fri Dec 23, 2011 1:07 am
What is the Source of this Question?? What level Question is this one ??

Is the Question out of Scope of GMAT???
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