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There are 8 teams in a certain league and each team plays each of the

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sarb

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Updated on: 26 Feb 2019, 04:29

Originally posted by sarb on 17 Jun 2012, 03:52.
Last edited by Bunuel on 26 Feb 2019, 04:29, edited 2 times in total.

Renamed the topic.

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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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Bunuel

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16 Nov 2012, 04:19

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Sachin9

Bunuel

sarb

The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).

Answer: C.

P.S. Please read and follow: https://gmatclub.com/forum/rules-for-pos ... 33935.html Pay attention ot the points #3 and #8.

I would like to learn about \(C^2_{8}=28\). Manhattan Book doesn't discuss this approach. They have anagram approach.

Well, the game is played by 2 teams. How many games are needed if there are 8 teams and each team plays each of the other teams exactly once? The number of games will be equal to the number of different pairs of 2 teams we can form out of 8 teams (one game per pair). How else?

Similar questions to practice:
https://gmatclub.com/forum/how-many-diag ... 01540.html
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https://gmatclub.com/forum/there-are-5-c ... 27235.html
https://gmatclub.com/forum/if-each-parti ... 42222.html

Hope it helps.

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12 Jan 2015, 14:46

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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:

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31 Jan 2013, 01:37

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it is not easy and is harder if we are on the test date.

there are 8 team
to take out 2 teams, IF ORDER MATTERS we have 8*7
but in fact order does not matter

8*7/2=28

princeton gmat book explain this point wonderfully.

Bunuel

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17 Jun 2012, 03:57

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sarb

The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).

Answer: C.

P.S. Please read and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention ot the points #3 and #8.

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14 Nov 2012, 06:23

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Bunuel

sarb

I would like to learn about \(C^2_{8}=28\). Manhattan Book doesn't discuss this approach. They have anagram approach.

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These type of problems can be solved with a simple diagram.

1. Draw a table consisting of 8 columns and 8 rows.
2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal.
3. The number should be 28.

I tried uploading the diagram but unsuccessful.

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16 May 2013, 01:13

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Bunuel

The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).

Hi Bunuel,

I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?

Thanks a lot.

Bunuel

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16 May 2013, 04:33

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mywaytomba

Bunuel

The total # of games played would be equal to the # of different pairs possible from 8 teams, which is \(C^2_{8}=28\).

Hi Bunuel,

I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula?

Thanks a lot.

Check combinatorics chapter of Math Book for theory: math-combinatorics-87345.html

Also check some questions on combinations to practice:
DS: search.php?search_id=tag&tag_id=31
PS: search.php?search_id=tag&tag_id=52

Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html (there are some about permutation too)

Hope it helps.

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26 Dec 2013, 23:04

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Lets assume the question asks There are 8 teams in a certain league and each team
plays each of the other teams exactly twice. If each
game is played by 2 teams, what is the total number
of games played?

Then is 28*2 the correct approach?

Bunuel

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27 Dec 2013, 03:08

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theGame001

Yes. Check here: 15-chess-players-take-part-in-a-tournament-every-player-55939.html and here: there-are-8-teams-in-a-certain-league-and-each-team-plays-132366.html

Similar questions to practice:
how-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html
if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html
10-business-executives-and-7-chairmen-meet-at-a-conference-126163.html
how-many-different-handshakes-are-possible-if-six-girls-129992.html
15-chess-players-take-part-in-a-tournament-every-player-55939.html
there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html
if-each-participant-of-a-chess-tournament-plays-exactly-one-142222.html
there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html
there-are-8-teams-in-a-certain-league-and-each-team-plays-132366.html
in-a-kickball-competition-of-9-teams-how-many-possible-ma-161846.html

Hope it helps.

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Updated on: 13 Jan 2015, 07:51

Originally posted by quentin.louviot on 26 Nov 2014, 02:14.
Last edited by quentin.louviot on 13 Jan 2015, 07:51, edited 2 times in total.

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SreeViji

Hi Bunnel,

I would also like to learn this approach. Can u help me?

Sree

Hey SreeViji,

I think i have something to help you.
The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2}

To understand that we just have to think that each of the 8 team plays against 7 other (8x7) but they play each team exactly once so we divide the total by 2.

We divide by 2 because "TEAM A VS TEAM B" is the same as "TEAM B VS TEAM A"

So we end up with \frac{8x7}{2}= 28

If you still have trouble with combination and permutation check out this website it's well done,
https://www.mathsisfun.com/combinatorics ... tions.html

hope it helps.

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19 May 2015, 07:31

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pranav123

Hi,

I don;t think we need to count the half spaces. with half space count is 36.
without half space - count: 28.

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onedayill

pranav123

Hi,

I don;t think we need to count the half spaces. with half space count is 36.
without half space - count: 28.

Dear onedayill

You're right!

The boxes along the diagonal (these are the boxes that contribute to half spaces) represent a team playing with itself. Since that is not possible, these boxes should not be included in the counting.

I noticed in the thread above that a few students had doubts about the expression 8C2. If any of the current students too have such a doubt, here's how this question could be solved visually:

There are 7 ways in which Team 1 can play with another team.
Similarly, there are 7 ways for each of the 8 teams to choose its playing opponent.

But it's easy to see that the red zone is essentially a duplication of the blue zone. For example, (Team 1 playing with Team 2) is the same case as (Team 2 playing with Team 1)

So, the correct answer will be: 8(that is, the number of teams)*7(that is, the number of ways in which each team can choose its playing opponent)/2 = 28

Hope this was useful!

Best Regards

Japinder

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25 May 2016, 22:25

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Attached is a visual that should help.

Attachments

Screen Shot 2016-05-25 at 9.41.57 PM.png [ 75.15 KiB | Viewed 298367 times ]

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sarb

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, as follows:

8C2 = 8! / [2! x (8-2)!]

(8 x 7 x 6!) / (2! x 6!)

(8 x 7)/2!

(8 x 7)/ 2

4 x 7 = 28

Answer C

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Updated on: 06 Dec 2018, 11:46

Originally posted by BrentGMATPrepNow on 27 Jun 2017, 06:43.
Last edited by BrentGMATPrepNow on 06 Dec 2018, 11:46, edited 1 time in total.

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sarb

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 28

Answer: C

Cheers,
Brent

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14 Oct 2017, 02:54

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sarb

One needs to always find n and r in permutation and combination problems. n is the total number from which you choose and r is how many you choose at a time which makes it one possibility. If the order of the entities chosen at a time, does not matter as in this case, it is nCr , otherwise it is nPr.

n and r in this problem are 8 and 2 respectively and the total number of games played is 8C2=28. Hence C.

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Bunuel

sarb

Bunuel lets say teams are as follows: A, B, C, D, E, F, G, H. So by using the simple combinatorics formula how do we exclude repeated teams ? I mean if A played with B - this is one game, and it could be also B WITH A ? yeah sounds a bit silly