Bunuel wrote:

Smita04 wrote:

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?

A. 15

B. 16

C. 28

D. 56

E. 64

# of different pairs possible from 8 teams is \(C^2_{8}=28\), since each pair plays twice between each other than total # of games is 2*28=56.

Answer: D.

I'm a little confused here -- why are we using the combination formula and NOT the permutation formula. We don't really care for these teams to be arranged alphabetically. Similar to if the letters are to be arranged alphabetically, meaning, ab, ac, ad, bc, bd, then we would use combination. But we don't care if team D plays B vs. team B playing team D. Since order is NOT important, wouldn't we use permutation.

There were a few similar problems:

1) How many 2 letters words can be made out of ABCD and in alphabetical order - 2C4 = 6

2) How many unique 4 letter words can be made from 10 letters but ABCDE and EDCBA are considered different = 10P4 = 10!/6!

Doesn't this question fall into the Permutation area?