Problem Solving

OG 12 Prob 121

Can anyone explain the prob in detail?

There are 8 teams lets say
a will play 7 games (with b,c,d,e,f,g,h)
b will play 6 games (with c d e f g h, game with "a" has already been counted)
c will play 5 games (with d e f g h, game with "a b" has already been counted)
d will play 4 games ( with e f g h, game with " a b c" has already been counted)
e will play 3 games ( with f g h, game with " a b c d" has already been counted)
f will play 2 games (with g h, game with " a b c d e" has already been counted)
g will play 1 game (with h, game with " a b c d e f" has already been counted)
h 0( every game has been counted played by h)

add all of them = 7+6+5+ 4+3+2+1 = 28(answer)

We can think of this as a pool of 8 teams, such that each matchup is a selection of 2 teams from the 8: the # of possible games = the number of possible pairing of two teams from the 8.

Essentially this is asking How many ways can we pick 2 teams from 8 available. This is the # of combinations of 2 from 8.

8!/6!2! = 28 ways.

-Patrick

another way to look at this:

set up a grid:
-----1--2--3--4--5--6--7--8
1
2
3
4
5
6
7
8

8 x 8 = 64 matchups
teams can't play themselves, so team 1 doesn't play team 1
therefore 56 games,
but team 1 playing team 2 is the same as team 2 playing team 1 and so on, so we need to divide by 2
=28