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

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?

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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.

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:

GMAT assassins aren't born, they're made,
Rich

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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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.

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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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?

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

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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.

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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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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Attached is a visual that should help.

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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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

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 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 but how do we exclude such repetition thanks!

8C2 gives the number of different unordered pairs possible from 8:
(A, B)
(A, C)
...
(B, H)
...
(G, H)

So, (A, B) is there only once (there is no (B, A) there)

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