A certain league

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A certain league

by alex.gellatly » Sun Sep 02, 2012 7:07 pm
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?

15
16
28
56
64

[spoiler]OK I know how to solve this problem, but I don't understand why MGMAT's slot method doesn't work here. I normally use the P or C formulas to solve these types of problems, but recently I've been trying the slot approach....am I missing something here..[/spoiler]

Thanks
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by pemdas » Sun Sep 02, 2012 8:46 pm
it's 8C2=28 or slot-wise one team 8 options and another 7 options. Adjusting for one game, cancelling ordered arrangement and leaving only combination 8*7/2!=28
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by Ian Stewart » Sun Sep 02, 2012 11:02 pm
alex.gellatly wrote: OK I know how to solve this problem, but I don't understand why MGMAT's slot method doesn't work here.
That method doesn't belong to any particular company. :)

The first question to ask yourself when counting is whether order matters. If it does, you can just use slots. If it doesn't, you can't just use slots - you'll need to account for the fact that order doesn't matter somehow. There are a few ways people learn to do this. You can use slots, and then at the end just divide by k! where k is your number of slots. Or you can use an nCk formula if you prefer, but there will always be an extra step if order doesn't matter.

In this question, order does not matter: "Chicago against New York" would be the same game as "New York against Chicago". So you can use slots as a starting point, but then because order doesn't matter, you will need to divide by 2! = 2 at the end. Here we'd have 8 choices for the first team in the game, and 7 choices for the second, so if order mattered, the answer would be 8*7. Since order doesn't matter, we need to divide by 2!, so we get 8*7/2! = 28 games.
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by talk2taran » Tue Sep 04, 2012 1:48 am
principle of counting can do the job :idea:

Match is played between 2 teams say _v/s _. Number of choices to fill the 1st dash are 8, which leaves us with 7 options for 2nd dash. so number of possible matches 8*7= 56
Now notice that in these 56 matches,say 2 teams (A and B) have played with each other twice. So divide by 2.

so final answer 56/2= 28 B-)
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