## Combination Problem

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### Combination Problem

by datonman » Wed Nov 12, 2014 12:31 pm

00:00

A

B

C

D

E

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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

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by Brent@GMATPrepNow » Wed Nov 12, 2014 12:36 pm
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
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).

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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by GMATGuruNY » Wed Nov 12, 2014 12:36 pm
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
Each game must be played by a UNIQUE PAIR of teams.
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.

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by GMATGuruNY » Wed Nov 12, 2014 12:56 pm
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.

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by Mathsbuddy » Fri Nov 14, 2014 6:18 am
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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by Brent@GMATPrepNow » Fri Nov 14, 2014 6:41 am
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
We can also solve this question using COMBINATIONS.

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/gmat-counting?id=789

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by gautamkumar » Sun Nov 16, 2014 8:58 am
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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by gautamkumar » Sun Nov 16, 2014 8:59 am
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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by Mathsbuddy » Wed Nov 19, 2014 7:06 am
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!

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by Rich.C@EMPOWERgmat.com » Thu Mar 22, 2018 11:39 am
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.

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

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by Jeff@TargetTestPrep » Mon Mar 26, 2018 4:51 pm
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
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:

8C2 = 8!/[2!(8 - 2)!] = 8!/(2!6!) = (8 x 7)/(2 x 1) = 56/2 = 28

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by BTGmoderatorRO » Thu Mar 29, 2018 10:40 am
Since each team plays each of the other at exactly once and each game is played by 2 teams , then
Total games played= $$8C_2$$
$$8C_2=\frac{8!}{\left(8-2\right)!2!}=\frac{8!}{6!2!}$$
$$=\frac{\left(8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1\right)}{\left(6\cdot5\cdot4\cdot3\cdot2\cdot1\cdot2\cdot1\right)}=\frac{\left(8\cdot7\right)}{2}=4\cdot7=28$$
$$Therefore,\ option\ C\ is\ correct$$

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