Combination Problem

Question

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

Brent@GMATPrepNow · Answer

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

As noted above, each game must be played by a UNIQUE PAIR of teams.
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.

GMATGuruNY · Answer

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

As noted above, each game must be played by a UNIQUE PAIR of teams.
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.

GMATGuruNY · Answer

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

As noted above, each game must be played by a UNIQUE PAIR of teams.
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.

Mathsbuddy · Answer

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.

Brent@GMATPrepNow · Answer

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

As noted above, each game must be played by a UNIQUE PAIR of teams.
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.

gautamkumar · Answer

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.

gautamkumar · Answer

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.

Mathsbuddy · Answer

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!

Rich.C@EMPOWERgmat.com · Answer

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

Contact Rich at [email protected]

Jeff@TargetTestPrep · Answer

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

As noted above, each game must be played by a UNIQUE PAIR of teams.
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.

BTGmoderatorRO · Answer

Since each team plays each of the other at exactly once and each game is played by 2 teams , then
Total games played= $$8C_2$$
$$8C_2=\frac{8!}{\left(8-2\right)!2!}=\frac{8!}{6!2!}$$
$$=\frac{\left(8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\right)}{\left(6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot2\cdot1\right)}=\frac{\left(8\cdot7\right)}{2}=4\cdot7=28$$
$$Therefore,\ option\ C\ is\ correct$$

Choose how you want to prepare

Get Ready for GMAT Test Day Faster with Live Online Classes

Target Test Prep EA OnDemand

Target Test Prep GMAT OnDemand

Combination Problem

GMAT/MBA Expert

GMAT/MBA Expert

GMAT/MBA Expert

Chicago Booth

GMAT Course Reviews

Admissions Consulting Reviews

FREE GMAT PREP RESOURCES

GMAT PREP DEAL TRACKER