In how many ways can nine students be divided into teams of $4$, $3$, and $2$ people?
The teams are distinguished by their sizes. Choosing who is on each team completely determines the teams.
There are $\binom{9}{4}$ ways to select four of the nine students to be on the team of four students, $\binom{5}{3}$ to select three of the five remaining students to be on the team with three students, and one way to form a team of two from the remaining two students. Hence, there are
$$\binom{9}{4}\binom{5}{3} = \frac{9!}{4!5!} \cdot \frac{5!}{3!2!} = \frac{9!}{4!3!2!}$$
ways to divide the nine students into three unlabeled teams.
If we had instead chosen the team of two, then the team of three from the remaining seven students, and then placed the remaining four students on the team of four, we could select the teams in
$$\binom{9}{2}\binom{7}{3} = \frac{9!}{2!7!} \cdot \frac{7!}{3!4!} = \frac{9!}{2!3!4!}$$
ways, in agreement with above.
Notice that labeling the team with four students team A, the team with three students team B, and the team with two students team C would not change our answer.
More care would be required if two or more of the groups had the same size.
Suppose our students are Amanda, Brenda, Claire, Dennis, Edward, Fiona, Gloria, Henry, and Ivan.
In how many ways can nine students be divided into three unlabeled teams of three people?
If we divide the nine students into teams of three, then the $3! = 6$ divisions
\begin{align*}
& \{Amanda, Brenda, Claire\}, \{Dennis, Edward, Fiona\}, \{Gloria, Henry, Ivan\}\\
& \{Amanda, Brenda, Claire\}, \{Gloria, Henry, Ivan\}, \{Dennis, Edward, Fiona\}\\
& \{Dennis, Edward, Fiona\}, \{Amanda, Brenda, Claire\}, \{Gloria, Henry, Ivan\}\\
& \{Dennis, Edward, Fiona\}, \{Gloria, Henry, Ivan\}, \{Amanda, Brenda, Claire\}\\
& \{Gloria, Henry, Ivan\}, \{Amanda, Brenda, Claire\}, \{Dennis, Edward, Fiona\}\\
& \{Gloria, Henry, Ivan\}, \{Dennis, Edward, Fiona\}, \{Amanda, Brenda, Claire\}
\end{align*}
are all equivalent since they result in the same three teams. Therefore, the number of ways of dividing the class into three unlabeled teams of three is
$$\frac{1}{3!}\binom{9}{3}\binom{6}{3} = \frac{1}{3!} \cdot \frac{9!}{3!3!3!}$$
We divide by $3!$ to account for the $3!$ orders in which we could select the same three teams of three.
In how many ways can the nine students be divided into three unlabeled teams of sizes $2$, $2$, and $5$?
Similarly, if the teams are not labeled and we divide the class into two teams of two and one team of five, the two divisions
\begin{align*}
\{Amanda, Brenda\}, \{Claire, Dennis\}, \{Edward, Fiona, George, Henry, Ivan\}\\
\{Claire, Dennis\}, \{Amanda, Brenda\}, \{Edward, Fiona, George, Henry, Ivan\}
\end{align*}
are equivalent since they result in the same three teams. Hence, the number of ways of dividing the nine students into two teams of two and one team of five if the teams are unlabeled is
$$\frac{1}{2!}\binom{9}{2}\binom{7}{2} = \frac{1}{2!} \cdot \frac{9!}{2!2!5!}$$
We divide by $2!$ to account for the $2!$ orders in which we could pick the same teams of size two.
If we had instead picked the team of five first, we would be left with four people. You might think that the two teams of two could be picked in $\binom{4}{2}$ ways, but this counts every team twice, once when we choose a team and once when we choose its complement. Alternatively, notice that if our team of five consists of Edward, Fiona, Gloria, Henry, and Ivan, the two teams of two are distinguished by who is paired with Amanda. There are three ways to do this:
\begin{align*}
\{Amanda, Brenda\}, \{Claire, Dennis\}\\
\{Amanda, Claire\}, \{Brenda, Dennis\}\\
\{Amanda, Dennis\}, \{Brenda, Claire\}
\end{align*}
Hence, the number of divisions of nine students into two teams of two and one team of five is
$$\binom{9}{5} \cdot 3 = \binom{9}{5} \cdot \frac{1}{2}\binom{4}{2} = \frac{1}{2}\binom{9}{5}\binom{4}{2} = \frac{1}{2!} \cdot \frac{9!}{2!2!5!}$$
Notice that the team of five is distinguished by its size, while the two teams of two are not. The teams of two can only be distinguished by who is on which team.
To summarize, teams of different sizes are distinguished by their sizes, so the order in which they are selected does not matter. If we have unlabeled teams of the same size, we have to divide by the number of orders in which we could pick the same teams.
Best Answer
For (a), as an earlier comment stated, you're making order matter by using a permutation. So, for example, you have $18$ choices for the first person on Team 1, $17$ choices for the second, and $16$ choices for the third. Just as you wrote, this makes $18\cdot 17\cdot 16$ choices. However, from those three people, we could have chosen the first person $3$ ways, the second $2$ ways, and the final person $1$ way. Therefore, $18\cdot 17\cdot 16$ overcounts the number of teams by a factor of $3\cdot 2 \cdot 1$. That means there are $\frac{18\cdot 17\cdot 16 }{3\cdot 2 \cdot 1}$ ways to choose the first team. This is equivalent to the binomial coefficient $\binom{18}{3}$, if you're familiar with combinations.
However, following this idea for each team will make your answer smaller, whereas your answer is too small to begin with. What you want to do is multiply those fractions rather than add them. Here we're determining one team and then determining another and then determining a third... Often, when an "and" appears as part of how you would complete a task (as above), that means you multiply. An "or", on the other hand, tells you to add; for example, if you wanted to make only one team, and you wanted to know how many ways there were to create a team of either three or four people from 18, you'd calculate
$$\frac{18\cdot 17\cdot 16 }{3\cdot 2 \cdot 1} + \frac{18\cdot 17\cdot 16\cdot 15}{4\cdot 3 \cdot 2 \cdot 1}$$