Let Teams are A B C D E F G H I J K L

Method - 1:

FIrst team will have 11 other teams to play a match with

This way each team will have 11 games to play with other teams

Total Matches = 11*12

Mistake: But here the matches have been counted twice e.g. A plays with B and then we count B's match with A likewise A plays with C and C plays with A

i.e. Total unique matches = 12*!1/2 = 66


Method - 2:

First team's matches = 11
Second team's matches = 10 (cause second has played a match with first)
Third team's matches = 9
and so on...

Total Matches = 11+10+9+8+7+6+5+4+3+2+1 = (1/2)*11*(11+1) = 66

Method - 3:

One match happens when two teams come together
i.e. total matches = total possible ways of forming groups of two teams

total ways of picking two teams out of 12 = 12C2 = 12! / (2!*10!) = 66

Answer: Option C
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total possible matches in tournament ; 12 c2 ; 66
IMO C

This ques reminds me of formula of number of handshakes between people.

- Say n be number of teams. n = 12
- Total number of matches: \(\frac{n(n-1)}{2 }=> \frac{12*11}{2} = 66\)

Option C!
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There are 12 teams.
If we ask each team, "How many teams did you play?" we'll find that each team played 11 teams, which gives us a total of 132games (since 12 x 11 = 132).

From here we must recognize that each game has been COUNTED TWICE.
For example, if Team A and Team B play a game, then Team A counts it as a game, AND Team B counts it as a game.

So, to account for the DUPLICATION, we'll divide 132 by 2 to get 66

Answer: C

Cheers,
Brent
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The number of games played is 12C2 = 12! / (2! x 10!) = (12 x 11) / 2 = 66.

Answer: C

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Every team plays 11 games so total number of games should be 12*11 but since in a match two teams are involved we half the obtained number.
Hence total matches = \(\frac{12*11}{2}\) = 66

Answer C.
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Asked: 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?

Number of games which will be played = 12C2 = 66

IMO D
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Total number of different matches = \(^{12}C_{2}\) = \(\frac{12*11}{2}\) = \(66\)
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For someone who doesnt know about combination formulae:

consider 4 teams A B C D

Listing down number of matches for these 4 teams as per the given conditions:

AvA - a team cannot play against itself
AvB
AvC
AvD
Number of matches: 3

BvA - Already counted above
BvB - a team cannot play against itself
BvC
BvD
Number of matches : 2

CvA - Already counted above
CvB - Already counted above
CvC - a team cannot play against itself
CvD
Number of matches : 1

DvA - Already counted above
DvB - Already counted above
DvC - Already counted above
DvD - a team cannot play against itself
Number of matches : 0

Total number of matches: 3+2+1+0=6
For 4 teams, number of matches as per given conditions is: 3+2+1 i.e (4-1)+(4-2)+(4-3)+(4-4)=3+2+1+0=6

You can calculate for any number of teams using this approach: Use Arithmetic Progression of common difference 1 for a higher number of teams as First term will be (Number of teams-1) and last term would be 1

For the given problem:
Total number of matches= 11+10+9+...+1 = {(11+1)*11}/2 = 66

Solution:

Number of games that can be played exactly one game with each of the other teams

= 12 C 3 (We select two teams of 12 and hence use combination)

=66 (option c)

Hope this helps

Devmitra Sen(GMAT Quant Expert)
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Answer: Option C

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