IanElliott wrote:
Quote:
There will be \(C^2_6=15\) games needed so that each team to play every other team exactly once (\(C^2_6=15\) is the # of ways we can pick two different teams to play each other).
[end quote]
Pardon me if the answer to my question has been posted already, but I do not understand the method some of you used to get 15 games:
Bunuel I am not familiar at all with the notation you used ( \(C^2_6=15\) )
Furthermore, some others here used 6!/(2!4!). Although I can see that it yields the right answer, I fail to understand the reasoning in how and why they chose those particular numbers. ie: 2! and 4! in the denominator. I guess I need to be enlightened about why factorials are a profitable tool here.
In my case, I used n(n-1)/2 which seemed to work okay. Is one method preferable to the other?
Hope my question was clear.
There are various ways to get a particular answer. People choose a particular method depending on how they think.
Your thought would have been something like this: There are 6 teams. The first team plays with every other team i.e. with 5 teams and goes away.
Now, the second plays with the rest of the 4 teams and goes away. Now the third team plays with the rest of the 3 teams and so on...
So total there are 5 + 4+ 3 + 2 + 1 and you used n(n-1)/2 here.
Another way to think about it is this: There are total 6 teams. Every pair of two teams played one match. Out of 6 teams, we will try to make as many distinct pairs of 2 teams each as possible. For this, we use 6C2 = 6!/2!*4!
There is absolutely nothing wrong with either one of the methods. But you must be able to think in nCr terms too because it comes in very handy in many questions. nCr = choose r items from n
For a background on combinations, check this post:
http://www.veritasprep.com/blog/2011/11 ... binations/