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Handshakes

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CIO
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Handshakes [#permalink] New post 14 Jul 2004, 06:16
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 0 sessions
There are 7 basketball teams at a tournament, each with 5 players. If each player will shake hands once with every other player not on his team, how many handshakes take place?

A) 324,632
B) 595
C) 525
D) 350
E) 120
Joined: 31 Dec 1969
Location: United States
Concentration: Marketing, Other
GMAT 1: 710 Q49 V38
GMAT 2: 660 Q V
GPA: 3.64
WE: Accounting (Accounting)
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Re: Handshakes [#permalink] New post 14 Jul 2004, 06:38
ian7777 wrote:
There are 7 basketball teams at a tournament, each with 5 players. If each player will shake hands once with every other player not on his team, how many handshakes take place?

A) 324,632
B) 595
C) 525
D) 350
E) 120


There will be 6*5 = 30 handshakes with persons who are not members of one's team. Given that we have total 7*5 = 35 players, we get 30*35 handshakes. But each handshake in this pool of h/s is counted twice. Thus, we have to discount and the answer is 30*35/2 = 525, C.
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 [#permalink] New post 14 Jul 2004, 07:21
Different approach:

7C2*5C1*5C1=21*5*5=525
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 [#permalink] New post 14 Jul 2004, 13:36
# of handshakes bween 2 teams = 5*5=25
# of comb to pick 2 teams = 7C2

Total comb = 25*7C2
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 [#permalink] New post 14 Jul 2004, 22:13
OA is C. I agree with emmanuel's way most of all. very straight forward and easy to comprehend: 35x30/2.
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 [#permalink] New post 15 Jul 2004, 13:50
Agreed, the combination method seems easy to mess up on, if you forget a factor to take into consideration. The dividing by two part is brilliant. Don't see that right off. Good job.
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Davefor MBA

  [#permalink] 15 Jul 2004, 13:50
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