Permutations/Combinations... : GMAT Problem Solving (PS)
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# Permutations/Combinations...

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06 Jan 2010, 05:50
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I got this question from a GMAT club in south korea...But i got a different answer...could someone explain the steps to solve this problem?

In a tournament of table tennis, a team consists of 2 members and the games between the two teams are performed once. If 8 persons are participated in the game, how many different ways are there that the games are performed with two teams?

A) 484 B) 548 C) 630 D) 842 E) 960. I worked out this way - using 8 people, we can form 28 different teams...(8*7/2!). these 28 teams can play different games in 28*27/2 ways = 378 games...But the OA here is (C). Can someone point out the flaw in my uderstanding?.. thank you.
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06 Jan 2010, 12:07
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raghuraoram wrote:
I got this question from a GMAT club in south korea...But i got a different answer...could someone explain the steps to solve this problem?

In a tournament of table tennis, a team consists of 2 members and the games between the two teams are performed once. If 8 persons are participated in the game, how many different ways are there that the games are performed with two teams?

A) 484 B) 548 C) 630 D) 842 E) 960. I worked out this way - using 8 people, we can form 28 different teams...(8*7/2!). these 28 teams can play different games in 28*27/2 ways = 378 games...But the OA here is (C). Can someone point out the flaw in my uderstanding?.. thank you.

There is a GMAT Club in South Korea?
Is it different from this one?
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06 Jan 2010, 19:02
raghuraoram wrote:
I got this question from a GMAT club in south korea...But i got a different answer...could someone explain the steps to solve this problem?

In a tournament of table tennis, a team consists of 2 members and the games between the two teams are performed once. If 8 persons are participated in the game, how many different ways are there that the games are performed with two teams?

A) 484 B) 548 C) 630 D) 842 E) 960. I worked out this way - using 8 people, we can form 28 different teams...(8*7/2!). these 28 teams can play different games in 28*27/2 ways = 378 games...But the OA here is (C). Can someone point out the flaw in my uderstanding?.. thank you.

But 8C2 only gives you 1 team. You still need to select members of the 2nd team (ie: you need 2 teams of 2 players each for each game).
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06 Jan 2010, 22:32
well...the club is not registered.. it is surely different from this one.

mrblack ..could you please elaborate. I am getting your point. but can you solve the problem with the steps?
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07 Jan 2010, 00:38
raghuraoram wrote:
I got this question from a GMAT club in south korea...But i got a different answer...could someone explain the steps to solve this problem?

In a tournament of table tennis, a team consists of 2 members and the games between the two teams are performed once. If 8 persons are participated in the game, how many different ways are there that the games are performed with two teams?

A) 484 B) 548 C) 630 D) 842 E) 960. I worked out this way - using 8 people, we can form 28 different teams...(8*7/2!). these 28 teams can play different games in 28*27/2 ways = 378 games...But the OA here is (C). Can someone point out the flaw in my uderstanding?.. thank you.

This is a complex question and I don't think the answer choices are correct.

The problem is that in a game, a person cannot be on two teams at the same time.

the number of 2 person teams from a group of 8 people is 8C2 = 28

If you want to visualize it:
let 8 people be represented by: a b c d e f g h

the teams they can form:
ab ac ad ae af ag ah
bc bd be bf bg bh
cd ce cf cg ch
de df dg dh
ef eg eh
fg fh
gh

28 teams

Each team can combine with one other team to form a game. (a person can only be on one team at a time):

ab can only play with teams that don't have a or b:
abcd abce abcf abcg abch abde abdf abdg abdh abef abeg abeh abfg abfh abgh

ac can only play with teams that don't have a or c:
acbd acbe acbf acbg acbh acde acdf acdg acdh acef aceg aceh acfg acfh acgh

ad can only play with teams that don't have a or d:

and so on. so for each team in the first row, they can play with 15 other teams ( 7 * 15)

However, since each game between two teams is performed only once, each team in a row cannot play with the previous row because they already played a game.

bc can only combine with teams that don't have b or c or a (because combinations with 'a' have already played):
bcde bcdf bcdg bcdh bcef bceg bceh bcfg bcfh bcgh

so each team in the 2nd row can combine with 10 other teams for a game (6 *10)

using that same logic, go down to the next row. cd can only play with teams that don't have c or d or a or b (because teams with a or b have already played):
cdef cdeg cdeh cdfh cdfh cdgh

so each team in the 3rd row can play with 6 other teams for a game (5 *6)

next row is 3 other teams. Check de:
defgh defh degh --> (4 * 3)

next row is 1 other team. check ef:
efgh (3 * 1)

the following rows will have nothing more to play with because any combination with them have already played, so the total number of games will be:

(7*15) + (6*10) + (5*6) +(4*3) + (3*1) = 210

I believe the answer should be 210.

(Although 210 is a factor of 630, I don't see how you can get to 630).
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Re: Permutations/Combinations...   [#permalink] 07 Jan 2010, 00:38
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# Permutations/Combinations...

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