If each of the 12 teams participating in a certain tournament plays ex

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

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!
_________________

total possible matches in tournament ; 12 c2 ; 66
IMO C

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

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

Answer: C

_________________

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.
_________________

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

Total number of different matches = \(^{12}C_{2}\) = \(\frac{12*11}{2}\) = \(66\)

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)

Answer: Option C

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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

_________________

In a tournament with 12 teams, each team plays exactly one game with each of the other teams. To determine the total number of games played, we can use the combination formula.

The number of ways to choose 2 teams from a set of 12 teams is given by the combination formula:

C(12, 2) = 12! / (2! * (12-2)!) = (12 * 11) / (2 * 1) = 66.

Therefore, there will be 66 games played in the tournament.

Hence, the correct answer is C. 66.