Question

Ten friends want to play a game.

They must be divided into three teams with three people in each team and one field judge.

In how many ways can they do it?

Answer 1

Find the how many ways can make a teams with three people in each team and one field judge:

It is the number of ways you can arrange the letters $\text{AAABBBCCCF}$, indicating that three players are on team $\text{A}$, three on team $\text{B}$, and three on team $\text{C}$, with one the field judge $\text{F}$.

Each different arrangement represents one of the ways you can form three teams of $3$ with a field judge.

That number of arrangements, and therefore the number of ways to divide the players as required,

$$\begin{gathered} \frac{10!}{3!\times 3!\times 3!\times 1!}=\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3!}{3!\times 3\times 2\times 1\times 3\times 2\times 1\times 1} \\ =\frac{10\times 3\times 4\times 7\times 2\times 5\times 2}{1} \\ =16800 \end{gathered}$$

Therefore, there are $16800$ ways can make a teams with three people in each team and one field judge.