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by ketkoag » Wed May 06, 2009 12:50 am
If each of the 12 teams participating in a certain tournament plays exactly one game
with each of the other teams, how many games will be played?
A. 144 B. 132 C. 66 D. 33 E. 23
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Re: teams

by Pranay » Wed May 06, 2009 1:31 am
ketkoag wrote:If each of the 12 teams participating in a certain tournament plays exactly one game
with each of the other teams, how many games will be played?
A. 144 B. 132 C. 66 D. 33 E. 23
--------------------------------------------------------------------------------------
No. of ways in which 12th team can play uniquely only once is 11
No. of ways in which 11th team can play uniquely only once is 10
.
.
.
No. of ways in which 2nd team can play uniquely only once is 1

=> Total no. of ways are 11+10+9+8+7+6+5+4+3+2+1 = 11*12/2 = 66 ways.


Thus, C
-------------------------------------------------------------------------------------

Please correct if wrong.
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Re: teams

by sureshbala » Wed May 06, 2009 1:52 am
ketkoag wrote:If each of the 12 teams participating in a certain tournament plays exactly one game
with each of the other teams, how many games will be played?
A. 144 B. 132 C. 66 D. 33 E. 23
Since each team plays with all the other teams exactly once, the total number of games = 12C2=66
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Re: teams

by Pranay » Wed May 06, 2009 3:30 am
I reached that answer but somehow went wrong while computing 12C2.

Anyways, Sureshbala thanks for the response as it has been a corrective measure for me .. :)
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