Question 1204436: A six-team hockey league has a schedule requiring each team to play each of the other teams 9 times. How many games will be played in total?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Let's consider the scenario that each team plays another team exactly once.
Label the 6 teams as A,B,C,D,E,F
Notation like AB means teams A and B play together.
Here are the outcomes for team A
AB, AC, AD, AE, AF
Here's team B's outcomes
BC, BD, BE, BF
We don't list BA because AB was already covered.
Keep this process going and we have this full schedule:
AB, AC, AD, AE, AF
BC, BD, BE, BF
CD, CE, CF
DE, DF
EF
There's a staircase pattern going on.
There are 5 items in row 1, 4 in row 2, etc.
5+4+3+2+1 = 15 games in the season if each team plays once against each other.
What's another way to arrive at 15?
Well consider two slots for the two teams chosen to play together.
We have 6 choices for the first slot and 5 for the next.
That's 6*5 = 30 permutations and 30/2 = 15 combinations.
We divide by 2 so we avoid double-counting. Something like AB is the same as BA.
Another way to arrive at the 15 is to use the nCr combination formula. Plug in n = 6 and r = 2.
And yet another way to get 15 is to look at Pascal's Triangle. Look at the row that starts with 1,6,... and count 2 spots over to land on 15.
As you can see, there are plenty of options here.
Since there are 15 different matchups when each team plays another team once, there must be 9*15 = 135 different games if each team plays another team 9 times.
Answer: 135 games
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