# Help me understand these examples?

1. May 28, 2012

### omoplata

Examples 5b and 5c from "A First Course in Probability, 7th edition" by Sheldon Ross. Images of the relevant pages are attached.

EXAMPLE 5b
Ten children are to be divided into an A team and a B team of 5 each. The A team
will play in one league and the B team in another. How many different divisions are
possible?

Solution.
There are
10!/(5!.5!) = 252 possible divisions.

EXAMPLE 5c
In order to play a game of basketball, 10 children at a playground divide themselves
into two teams of 5 each. How many different divisions are possible?

Solution.
Note that this example is different from Example 5b because now the order
of the two teams is irrelevant. That is, there is no A and B team, but just a division
consisting of 2 groups of 5 each. Hence, the desired answer is
10!/(5!.5!.2!) = 126

I disagree.

I think for Example 5b the answer should be (10!/(5!.5!))*(10!/(5!.5!)). Because the teams A and B play in two different leagues, the same person can play in both teams. So there are 10C5 = 10!/(5!.5!) ways of choosing 5 people for team A, and the same number of ways for choosing 5 people for team B.

For example 5c, I think the answer should be 10C5 = 10!/(5!.5!). The two teams seem to be playing against each other, so no one can play both teams. So the number of ways of choosing the teams is just the number of way of choosing 5 people from 10. The remaining 5 will make up the other team.

But maybe the author of the book and I are not thinking along the same lines. Can someone point out my error and show me where I went wrong?

Thanks.

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2. May 28, 2012

### omoplata

The images look too small. If anyone wants a better image let me know. But the relevant problems are fully stated in the post.

3. May 28, 2012

### laughingebony

Regarding the first question:

If I gave you 10 apples and asked you to divide them into 2 groups of 5 apples each, would you put any of the apples in both groups simultaneously?

Regarding the second question:

Even though the author says there's no A team and no B team, I think it helps in understanding the solution to suppose that there is an A team and a B team. $C^{10}_5$ is then the number of ways to choose 5 people specifically for team A, while the other 5 go to team B. But this approach treats the following, for example, as separate events:

Team A -- 1,2,3,4,5
Team B -- 6,7,8,9,10

and

Team A -- 6,7,8,9,10
Team B -- 1,2,3,4,5.

But in the context of the problem, there's no difference between these two events, since the two teams will be playing against each other. We need to divide out the number of ways of permuting the teams, which is 2!.

4. May 28, 2012

Thanks.