# How many ways are there to split $10$ people into two groups of $5$? [duplicate]

In the example below I get the first part. May I know why they're dividing by 2 for the second part? I'm asking because I feel the answer for second part should also be $\binom{10}{5}$.

In the first part simply by choosing 5 people, we're already splitting the squad into two teams. So I don't really see a difference between parts 1 & 2. Help?

# Example

There are ten people in a basketball squad. Find how many ways:

1. the starting five can be chosen from the squad
2. the squad can be split into two teams of five.

## Solution

1. There are $\binom{10}{5} = \frac{10\times9\times8\times7\times6}{5\times4\times3\times2\times1} = 252$ ways of chosing the starting five.
2. The number of ways of dividing the squad into two teams of five is $\frac{252}{2} = 126$.
• in the first case, the order of team does matter, in the second one it does not Commented Jul 19, 2018 at 13:00

In the first one, there is a difference between the two resulting teams. One is playing, the other is sitting on the sidelines. So swapping the two teams around makes for a different choice of starting five.

In the second case, there is no difference. The only thing that matters is who is on team with who, and not which team is team A and what team is team B. So swapping the two teams around still gives the same choice of two teams.

You can also go the other way: Choose a starting five by first dividing into two teams (which can be done in some number of ways), then choosing which of the two teams gets to be athe starting five. You have two options for starting five for each division into two teams. Thus you have to multiply by 2 to get from one to the other.

Note that when we choose $\binom{10}{5}$ we are counting twice the same teams, as for example

$$\{ A,B,C,D,E \} , \{ G,H,I,L,M \} \equiv \{ G,H,I,L,M \},\{A,B,C,D,E\}$$

therefore the number of ways to obtain 2 teams is $\frac12 \binom{10}{5}$.

Observe that the choice of choosing five people and the other five are counted separately in the first case, but yield the same separation into two groups in the second. That's why there is division by $2$.

In the first case each choice is about selecting 5 out of ten, let's say from $\{A,B,C,D,E,F,G,H,I,J\}$. Note that selecting $\{A,B,C,D,E\}$ id different, than selecting $\{F,G,H,I,J\}$.

In the second case you select 5, but the other 5 are also forming a team. Thus your selection is for example $\{\{A,B,C,D,E\},\{F,G,H,I,J\}\}$. This selection is the same, as selecting $\{\{F,G,H,I,J\},\{A,B,C,D,E\}\}$. Thus in the way, that you form two teams by selecting 5 people to one and rest to another leads to duplication of selections. Therefore you need to divide it by the number of permutations of the teams, ie $2!$.

The second part of your problem will be $\binom{10}{5}$ if the question is of the following form:

How many ways to split $10$ people into two groups of $5$ people so that the first group is Team A and second group is Team B?

Here, unlike the original question, we restrict ourselves to a particular arrangement between the groups. If we don't have such a restriction as in the original post, we will have to divide by the number of possibilities to order the groups which is $2!$. Then the answer would become $$\frac{\binom{10}{5}}{2!}=126.$$

Imagine that you're one of the $10$ players. In both cases you need to determine who your $4$ teammates will be, out of the other $9$ players. This can be done in $\binom{9}{4} = 126$ ways.

Now, if the two teams are treated differently, like one team gets to play while the other team has to sit on the bench, then (once you have determined your teammates) you still have to determine which of the $2$ possible assignments your team gets. The two outcomes are rather different for you, so we have to count them differently. There are thus $2\binom{9}{4} = 252$ possible outcomes.

On the other hand, if there is no difference between teams, like if the two teams are just going to play each other, then they are not treated differently. It is not really possible to distinguish between the following cases:

(I) You, Alice, Bob, Charlie and Danielle are going to play against the other five players

or

(II) The other five players are going to play against you, Alice, Bob, Charlie and Danielle.

Both (I) and (II) describe the same event. That event should not be counted twice. So there are just $\binom{9}{4} = 126$ possible outcomes here.

This is from Sheldon Ross's book a First Course in Probability, I remember that example confusing me. Sometimes it helped me to think of it this way - assume that the first 5 picked will be on the red jersey team and the remaining will be on the yellow jersey team for option 1. Then a team of A,B,C,D,E in red jerseys is different than a team A,B,C,D,E in yellow jerseys, so they are both counted. In option 2, this is irrelevant so we divide by 2 to take out this distinction.