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

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.

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.

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
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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.

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.

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.

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

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

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

There's a formula for questions such as this, similar to the formula for calculating the number of handshakes: \(\frac{n(n-1)}{2}\)

\(\frac{8(8-1)}{2}\)= 28

C

Hi Dinesh654,

The prompt tells us that each Team plays each of the other times ONCE. In your calculation though, you have them playing TWICE. Here's why.

We can start by referring to the Teams as A, B, C, D, E, F, G and H.

When Team A plays Team B, we have 1 game - and we would end up with 7 total games involving Team A (one per opponent). We have to make sure that Team A does NOT play any of those other teams again.

If you next take Team B - and then choose Team A as an opponent - then you've created a 2nd game between them, which is not allowed. Thus, under these conditions, you cannot simply multiply 8c1 and 7c1 - since this would actually create TWO games for each pairing.

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

Contact Rich at: Rich.C@empowergmat.com

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)

Similar questions to practice:
https://gmatclub.com/forum/how-many-diag ... 01540.html
https://gmatclub.com/forum/if-10-persons ... 10622.html
https://gmatclub.com/forum/10-business-e ... 26163.html
https://gmatclub.com/forum/how-many-diff ... 29992.html
https://gmatclub.com/forum/15-chess-play ... 55939.html
https://gmatclub.com/forum/there-are-5-c ... 27235.html
https://gmatclub.com/forum/if-each-parti ... 42222.html
https://gmatclub.com/forum/there-are-8-t ... 34582.html
https://gmatclub.com/forum/there-are-8-t ... 32366.html
https://gmatclub.com/forum/in-a-kickball ... 61846.html
https://gmatclub.com/forum/there-are-8-t ... 34582.html

EMPOWERgmatRichC
Thanks for the nice explanation.
In your explanation, zero (in the highlighted part) indicates that team H payed with team H. We should not count H vs H here, actually. Isn't it?
Thanks__

That question is not the same:

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?

Hi Asad,

The "0" at the end represents the additional number of games that Team H would have to play. Since Team H already played all of the other teams (and those games have been 'counted' already), there's nothing else to add to the total.

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

Hi waihoe520,

The prompt leaves the exact nature of what's being 'played' as an unknown, but if it helps, then you might assume that the 8 teams are probably playing some type of sport (such as basketball or baseball), although that's not really relevant to the math involved. There will be some type of competition in which each team plays each other team ONCE.

If we call the 8 teams: A, B, C, D, E, F, G and H

Then we know that Team A plays B, C, D, E, F, G, and H.... which is 7 games total

Since Team B has ALREADY played Team A, those teams CANNOT play again... thus, Team B plays C, D, E, F, G and H = 6 more games

Team C has ALREADY played Team A and Team B, so it would play D through H = 5 more games

Etc.

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