How to find the number of possible outcomes of 10 games between 20 teams? Hi I am looking for an equation to find possible combinations in a non repeating format with a twist.  Here is the example:  There are 10 games between 20 teams.  I have to chose 5 winners but obviously when one slot is filled it would eliminate an additional option with itself because you can't have two winners from one game so how would you factor that in to the equation $\frac{n!}{r!(n-r)!}$ 
Let me know if you need me to explain it better.
 A: Let me see if I get this straight.
There are $20$ teams arranged in $10$ pairs. If team $i$ is paired with team $j$, then the two teams are playing a game with each other, and clearly, only one will win. 
You want to choose $5$ out of the $20$ original teams in the tournament that will win. 
You are interested in picking $5$ teams in the beginning of the tournament that you believe will win and this is your "roster". 
In the end of the day, since you only have to pick $5$ winning teams, you will focus on $5$ out of the $10$ games. There are ${10 \choose 5}$ ways you can choose which games you will focus on.
Once the set of $5$ games is fixed, there are $2^{5}$ possible outcomes for this set of $5$ games and you can choose one of them to determine your "winning" roster. 
So you can build your roster in 
$$
{10 \choose 5}2^{5}
$$
ways.
A: First you choose $10$ teams out of $20$ to play each other, and then you multiply that with the number of ways you can choose $5$ teams out of $10$ as winners (the actual games are really irrelevant when you are looking at all the possible outcomes).
A: If I am understanding your question correctly, you wish to take $2n$ teams, pair them up, and from among the winners pick some subset of the winners for some special reason such as playing in a charity event.  By means of a small example: 2 games, 4 teams ($ABCD$).  I expect you are interested in not only which two teams are the winners, but also who it is they played against but order in which the games are listed is unimportant (i.e., it didn't matter if team $A$ played team $C$ and the game was aired on espn5 or if it was aired on espn72).  Momentarily ignoring the current requirement to pick a subset of the winners for this example, we have.
In the format WinLose-WinLose the twelve possibilities are $AB-CD,~BA-CD,~AB-DC,~BA-DC,~~ AC-BD,~CA-BD,~AC-DB,~CA-DB,~~AD-BC,~DA-BC,~AD-CB,~DA-CB$.  (Again, I am considering $AB-CD$ to be the same as $CD-AB$)
With the inclusion of picking which subset of the winning teams to play for some charity event afterwards, you can modify the previous statement by bolding the selections:  I.e., $\textbf{A}B-CD$, $AB-\textbf{C}D$, and $\textbf{A}B-DC$ are all considered different outcomes.

This can be broken up into a multiplication principle problem as follows:
With $2n$ teams playing $n$ games, to select $r$ teams from among the winners
$\bullet$ step1: pick how to split up total group into ordered groups of two to play against eachother: $\frac{(2n)!}{2^n}$
$\bullet$ step2: divide by symmetry for the order in which matchups were picked (i.e., forget which tv channel they were aired on), $\frac{1}{n!}$
$\bullet$ step3: pick who won in each game, $2^n$
$\bullet$ step4: from the $n$ winners, pick $r$ of them to play for the charity event. $\binom{n}{r}$
For a final formula of $\frac{(2n)!}{2^n}\cdot\frac{1}{n!}\cdot2^n\cdot\frac{n!}{r!(n-r)!} = \frac{(2n)!}{r!(n-r)!}$

Checking the formula: had I used it for my example above and "picked zero" for the charity event: num of games = $n=2$, num of teams = $2n = 4$, num chosen = $r = 0$
$\frac{4!}{0!(2-0)!} = 4\cdot3 = 12$
Had I picked one winner for the charity event, there were two ways of choosing, so by the brute force method above I get 24 possibilities, and by the formula I get $\frac{4!}{1!(2-1)!} = 4\cdot3\cdot2 = 24$

For your specific example, we have $2n=20$, $n=10$, $r=5$ for a total of $\frac{20!}{5!5!}$ number of ways.
In the case the mention of the number 5 was unintentional and you meant only to choose the 10 winners without wanting to worry about a charity event afterwards, you can modify my formula by setting $r=0$ or $r=n$

Now with a formula, a completely different approach comes to mind:  Take the $2n$ objects and arrange them in a row ($2n!$ number of ways to do this).  Consider a single case of a slightly larger example with $n=4$ and $r=2$:  $ABCDEFGH$.  Place a separator after $r$ entries, and another after $n-r$ entries.  $AB-CD-EFGH$.  The $i$th team appearing in the list (with $i\leq n$) plays against the $n+i$th team in the list.  The group to the left of the first separator are the special winners, the group inbetween the two separators are the regular winners, and to the right are the losers.  Since we overcounted and consider $AB-CD-EFGH$ the same as $BA-CD-FEGH$ and $AB-DC-EFHG$, we need to divide by symmetry.  When moving one of the winners around in its area, we move the matching loser in the same way to keep it "the same" outcome, so we divide by $r!$ and $(n-r)!$ for a final answer again of $\frac{(2n)!}{r!(n-r)!}$
A: I'm doing a bad job of writing this out but I will try one more time.  Ok so a regular problem you have 20 teams and you want to see how many non repeating combinations of 5 you can make so you would simply do 20! over 5!(20-5)! and the answer would be 15,504.  Now what is different about what I am asking from this problem above is that once I chose my first team the opponent is eliminated from the rest of that combination so I were trying to figure out a 5 digit non repeating permutation from the 20 it would go 20x18x16x14x12 instead of 20x19x18x17x16.  I still don't think you guys are going to understand what I'm trying to say but oh well I will just try to figure it out myself I'll get it eventually.
