Question 1192101: In a volleyball game, each team must play with all the other teams. How many games will there be?
Found 2 solutions by math_helper, Theo:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website!
You didn't specify how many teams there are, and the answer depends on the number of teams, N.
Looking at TOTAL games played here, by all teams:
For N = 2, obviously 1 game will be played
For N = 3, there will be 3 games (A,B,C: unique games: A plays B, B plays C, A plays C)
For N = 4, there will be 6 games (A,B,C,D: unique games: A plays B,C,D; B plays C,D, C plays D)
... For the general case, there are  games played in TOTAL.
[ While each INDIVIDUAL team plays N-1 games, there is overcounting if we just multiply N-1 by N. See for example the N=4 case where I omitted "B plays A" because that is already counted in "A plays B". Hope this makes sense. ]
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! i believe it will be a combination formula that allows you to find the answer to this.
the formula is c(n,x).
n is the number of teams
x is the number of 2 team games.
c(n,x) = n! / (x! * (n-x)!)
the following excerpt from excel shows you how many games need to be played.
n x c(n,x)
2 2 1
3 2 3
4 2 6
5 2 10
6 2 15
i'll use A, B, C, ..... to identify each team.
when n = 2 and x = 2, you get 1 game that needs to be played.
that game is played between A and B, which will be shown as the set AB.
when n = 3, 3 games need to be played.
they will be between:
AB
AC
BC
what this means is:
A plays with B
B plays with C
B plays with C
if you look at it from the point of the team playing the other teams, you get:
A plays with B and C
B plays with A and C
C plays with A and B
when A plays with B, B is automatically playing with A.
that's AB and BA equals the same game.
that's what the combination formula does.
it tells you how many sets of 2 you can get from a set of n when order doesn't matter.
order doesn't matter means that AB and BA are the same set.
if order matters, then AB and BA would be different sets.
when n = 4, the number of games that have to be played so each team plays with each other team would be (4,2) = 6.
that works out to be:
AB
AC
AD
BC
BD
CD
if you look at it from the standpoint of A, A plays with B, C, and D.
if you look at it from the standpoint of B, B plays with A. C, and D.
if you look at it from the standpoint of C, C plays with A, B, and D.
if you look at it from the standpoint of D, D plays with A, B, and C.
every team has played with every other team exactly 1 time.
this occurs in 4 games.
the formula used is the combination formula.
it tells you how many sets of x you can get from a set of n when order doesn't matter.
when x = 2, then it's telling you how many sets of 2 you can get from a set of n when order doesn't matter.
once again, order doesn't matter means that AB and BA are considered part of the same set and are only shown as either AB or BA, but not both.
to show you the difference i will do C(4,2) and P(4,2).
C(n,x) formula is n! / (x! * (n-x)!)
P(n,x) formula is n! / (n-x)!)
C(4,2) = 6
P(4,2) = 12
the number of combinations of sets of 2 from 4 are:
AB
AC
AD
BC
BD
CD
that's a total of 6.
the number of permutations of sets of 2 from 4 are:
AB, BA
AC, CA
AD. DA
BC, CB
BD, DB
CD, DC
that's 12.
where AB and BA were considered part of the same set in C(4,2), AB and BA were considered part of different sets in P(4,2).
with the combination formula, order doesn't matter.
with the permutation formula, order does matter.
the number of games that must be played where each team plays each other team uses the combination formula.
here's a reference that might help you to understand the difference.
https://www.mathsisfun.com/combinatorics/combinations-permutations.html
note that the combination and permutation formulas assume selections without replacement.
note also that:
C(n,x) = P(n,x) / x!
P(n,x) = C(n,x) * x!
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