# Number of ways to divide $10$ children into two teams of $5$?

Consider the following two examples from A First Course in Probability by Sheldon Ross,

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 by the Book itself There are $$\frac{10!}{5!5!} = 252$$ possible divisions.

My Solution and Reasoning We first choose five children which can be done in $$C(10, 5)$$ possible ways. Then we decide which team(either A or B) to assign these children to, which can be done in $$2!$$ ways. Finally we assign the remaining five children to the remaining team, depending on the second level(if the first five children were assigned to A, then the remaining five should be assigned to B, and if the first five were assigned to B, the remaining to A). Therefore this level has only $$1$$ possible way, and hence the final answer is $$C(10, 5) \times 2! \times 1 = 252 \times 2 \times 1 = 504$$.

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 by the Book itself 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

$$\frac{10!/(5!5!)}{2!} = 126$$

My Solution and Reasoning First we choose 5 children for one team which can be done is $$C(10, 5)$$ ways and then we put the remaining children in the other team, which is done in $$1$$ way. Thus the answer is $$C(10, 5) \times 1 = 252$$, which can also be obtained by dividing the answer to Example 5b by 2, or in other words, removing the order of the groups in Example 5b.

There seems to be a conceptual misunderstanding in my way of thinking, but what is it?

PARTICULAR PROBLEM

For your particular problem, for labeled teams

• label the children $$0\; through\;$$9
• choose $$5$$ for team $$A$$, say $$13579$$, team $$B$$ automatically $$02468$$
• This is different from $$02468$$ in Team $$A$$ and $$13579$$ for team $$B$$, so total $$\binom{10}5$$ choices

But if teams are unlabeled $$[13579-02468] \equiv [02468-13579]$$
hence division by $$2$$

GENERAL GUIDANCE

It is important to have a clear concept over combinatorics of teams depending on whether they are labeled or unlabeled

To take a more complex example for greater clarity, suppose you are to choose $$4$$ teams of $$3$$ each from $$12$$ players

• If the teams are labeled, eg Horses, Greyhounds , Panthers, etc
the answer is $$\binom{12}3\binom93\binom63\binom33$$
which can be variously written as $$\binom{12}{3,3,3,3}\; or\; \dfrac{12!}{3!3!3!3!}$$

• If the teams are unlabeled, we shall have to divide by $$4!$$, to remove permutations between identical teams

• An important point to note is that unlabeled teams in effect become labeled if sizes differ, or composition differs (eg boys' team or girls' team)

Note that in both casees your answer is doubled from the book's solution so you should question whether you really ought to multiply by $$2!=2$$.

Let's consider a simpler case of 4 children $$\{1,2,3,4\}$$ and the teams are of size 2. Then all possible options for Team A are $$\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}$$ giving a total of $$\frac{4!}{2!2!}=6$$ options. What you are doing is counting each option twice; Say when you count $$\{1,2\}$$, you count once for this being assigned to $$A$$ and once for $$B$$. However, note that the event of $$\{1,2\}$$ belonging to $$B$$ is exactly the event that $$\{3,4\}$$ belongs to $$A$$ which we already count. Hence we should not multiply by 2.