Handshakes : DS Archive
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# Handshakes

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CIO
Joined: 09 Mar 2003
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14 Jul 2004, 06:16
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Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 2 sessions

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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: Russian Federation
Concentration: Technology, Entrepreneurship
GMAT 1: 710 Q49 V0
GMAT 2: 700 Q V
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 7: Q42 V44
GMAT 8: Q42 V44
GMAT 9: 740 Q49 V42
GMAT 10: 740 Q V
GMAT 11: 500 Q47 V33
GMAT 12: 670 Q V
WE: Engineering (Manufacturing)
Followers: 0

Kudos [?]: 202 [0], given: 102287

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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.
Intern
Joined: 07 Jul 2004
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14 Jul 2004, 07:21
Different approach:

7C2*5C1*5C1=21*5*5=525
Director
Joined: 05 May 2004
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Location: San Jose, CA
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14 Jul 2004, 13:36
# of handshakes bween 2 teams = 5*5=25
# of comb to pick 2 teams = 7C2

Total comb = 25*7C2
CIO
Joined: 09 Mar 2003
Posts: 463
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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.
Intern
Joined: 07 Jun 2004
Posts: 29
Location: Sunnyvale
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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.
_________________

Davefor MBA

15 Jul 2004, 13:50
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# Handshakes

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