combination tricks: each team plays each of the other teams exactly once

Ques: 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 three methods to solve such ques.

1st) The total # of games played would be equal to the # of different pairs possible from 8 teams, which is C28=28$C_8^2=28$.
Formula will be 


n!/(n-r)!*r!

=8!2!6!=4∗7=28


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 there are n teams need to play exactly once , then they play with (n-1) teams but as they are playing together then, n(n-1)/2, which means nC2

So 8*7/2 =28.

2nd) 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.
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.

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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


3rd) I also used a table to do this, like that:

1_2_3_4_5_6_7_8
1_1_1_1_1_1_1_1
2_2_2_2_2_2_2_2
3_3_3_3_3_3_3_3
4_4_4_4_4_4_4_4
5_5_5_5_5_5_5_5
6_6_6_6_6_6_6_6
7_7_7_7_7_7_7_7
8_8_8_8_8_8_8_8

Then you delete the same team pairs: e.g. 1-1, 2-2, 3-3 and then 2-1 (because you have 1-2), 3-2 (because you have 2-3). After you cross out the first 2 columns you then see that you cross out everything from the diagonal and below. The remaining is 28.

However, the 8!/2!*6! the approach is better because if you have many numbers the table will take forever to draw. In case there is sth similar though and your brain gets stuck

Hope now we learn possible tricks to solve such questions