There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
(A)15
(B)16
(C)28
(D)56
(E)64
Combination Problem
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 Brent@GMATPrepNow
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There are 8 teams. If we ask each team, "How many teams did you play?" we'll find that each team played 7 teams, which gives us a total of 56 games (since 8 x 7 = 56).datonman wrote:There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
(A)15
(B)16
(C)28
(D)56
(E)64
From here we need to 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 ALSO counts it as a game.
So, to account for the DUPLICATION, we'll divide 56 by 2 to get [spoiler]28 (C)[/spoiler]

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Each game must be played by a UNIQUE PAIR of teams.datonman wrote:There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
(A)15
(B)16
(C)28
(D)56
(E)64
Thus, the total number of games = the total number of unique pairs that can be formed from the 8 teams.
Number of options for the first team selected = 8. (Any of the 8 teams.)
Number of options for the second team selected = 7. (Any of the 7 remaining teams.)
To combine these options, we multiply:
8*7.
Since the ORDER of the two teams doesn't matter  AB is the same pair as BA  we divide by the number of ways the two teams can be arranged (2!):
(8*7)/(2*1) = 28.
The correct answer is C.
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 GMATGuruNY
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As noted above, each game must be played by a UNIQUE PAIR of teams.datonman wrote:There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
(A)15
(B)16
(C)28
(D)56
(E)64
An alternate approach is to WRITE OUT all of the possible pairings.
Let the 8 teams be A, B, C, D, E, F, G and H.
Pairs with A:
AB, AC, AD, AE, AF, AG, AH.
7 options.
Pairs with B (excluding AB, which has already been counted):
BC, BD, BE, BF, BG, BH.
6 options.
Pairs with C (excluding AC and BC, which have already been counted):
CD, CE, CF, CG, CH.
5 options.
Notice the PATTERN.
Options for A = 7.
Options for B = 6.
Options for C = 5.
With each successive letter, the number of options decreases by 1.
Thus:
Options for D = 4.
Options for E = 3.
Options for F = 2.
Options for G = 1.
Adding together the options above, we get:
Total number of games = 7+6+5+4+3+2+1 = 28.
The correct answer is C.
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Another way to solve this quickly is to sketch 8 dots in a circle and then connect each dot in turn to every other dot. There will be 28 lines. Answer = C.
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We can also solve this question using COMBINATIONS.datonman wrote:There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
(A)15
(B)16
(C)28
(D)56
(E)64
It takes two teams to play a game.
So, in how many different ways can we select 2 teams from 8 teams?
Since the order in which we select the teams doesn't matter, we can use combinations.
We can select 2 teams from 8 teams in 8C2 ways
8C2 = 28
If anyone is interested, we have a free video on calculating combinations (like 8C2) in your head: https://www.gmatprepnow.com/module/gmatcounting?id=789
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Somehow I have known the answer practically since last 22 years. 1992 Cricket world cup was played on all play all basis and my father explained it to me with the approach of A plays B the B plays A, but I wanted to know in terms of Combination.

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Somehow I have known the answer practically since last 22 years. 1992 Cricket world cup was played on all play all basis and my father explained it to me with the approach of A plays B the B plays A, but I wanted to know in terms of Combination.

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Another quick solution is to consider the 8 teams as 8 integers used in a 2 digit number.
In base 8, the highest value is 77 (base 8) = 7x8 + 8 = 64 (i.e there are 64 combinations of pairs of 8 integers)
Then we need to exclude double digits (8 of them) to get 56
and halve the answer to remove palindromic values (AB = BA), to get 56/2 = 28
Job done!
In base 8, the highest value is 77 (base 8) = 7x8 + 8 = 64 (i.e there are 64 combinations of pairs of 8 integers)
Then we need to exclude double digits (8 of them) to get 56
and halve the answer to remove palindromic values (AB = BA), to get 56/2 = 28
Job done!
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Hi All,
Using the Combination Formula IS one way to approach these types of questions, but it's not the only way. Sometimes the easiest way to get to a solution on Test Day is to just draw a little picture and keep track of the possibilities....
Let's call the 8 teams: ABCD EFGH
We're told that each team plays each other team JUST ONCE.
Start with team A....
A plays BCD EFGH = 7 games total
Team B has ALREADY played team A, so those teams CANNOT play again...
B plays CD EFGH = 6 more games
Team C has ALREADY played teams A and B, so the following games are left...
C plays D EFGH = 5 more games
At this point, you should notice a pattern: the additional number of games played is reduced by 1 each time. So what we really have is:
7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 28 games played
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
Using the Combination Formula IS one way to approach these types of questions, but it's not the only way. Sometimes the easiest way to get to a solution on Test Day is to just draw a little picture and keep track of the possibilities....
Let's call the 8 teams: ABCD EFGH
We're told that each team plays each other team JUST ONCE.
Start with team A....
A plays BCD EFGH = 7 games total
Team B has ALREADY played team A, so those teams CANNOT play again...
B plays CD EFGH = 6 more games
Team C has ALREADY played teams A and B, so the following games are left...
C plays D EFGH = 5 more games
At this point, you should notice a pattern: the additional number of games played is reduced by 1 each time. So what we really have is:
7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 28 games played
Final Answer: C
GMAT assassins aren't born, they're made,
Rich
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We are given that there are 8 teams in a league and that each game is played by 2 teams. Note that each team does not play itself and the order of pairing each team with its opponent doesn't matter. [For example, the pairing of (Team A vs. Team B) is identical to the pairing of (Team B vs. Team A).] The situation can therefore be solved by finding the number of combinations of 8 items taken 2 at a time, or 8C2:datonman wrote:There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?
(A)15
(B)16
(C)28
(D)56
(E)64
8C2 = 8!/[2!(8  2)!] = 8!/(2!6!) = (8 x 7)/(2 x 1) = 56/2 = 28
Answer: C
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Since each team plays each of the other at exactly once and each game is played by 2 teams , then
Total games played= $$8C_2$$
$$8C_2=\frac{8!}{\left(82\right)!2!}=\frac{8!}{6!2!}$$
$$=\frac{\left(8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\right)}{\left(6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot2\cdot1\right)}=\frac{\left(8\cdot7\right)}{2}=4\cdot7=28$$
$$Therefore,\ option\ C\ is\ correct$$
Total games played= $$8C_2$$
$$8C_2=\frac{8!}{\left(82\right)!2!}=\frac{8!}{6!2!}$$
$$=\frac{\left(8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\right)}{\left(6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot2\cdot1\right)}=\frac{\left(8\cdot7\right)}{2}=4\cdot7=28$$
$$Therefore,\ option\ C\ is\ correct$$