How many possible combinations of teams to tasks (GRE math subject test) The question is:
A teacher is assigning 6 students to one of three tasks. She will assign students in teams of at least one student, and all students will be assigned to teams. If each task will have exactly one team assigned to it, then which of the following are possible combinations of teams to tasks?
I. 90 . II. 60 . III. 45
(A) I only
(B) I and II only
(C) I and III only
(D) II and III only 
(E) I, II, and III
The answer in the book gives B, and explanations are attached below as a screenshot. However, personally I only want to select A.
For the case where each team has 2 people, then I agree that the total number of combinations is: $$\binom{6}{2}\binom{4}{2}\binom{2}{2}$$, which gives 90.
However, for the case where these three teams have 3, 2, and 1 people, I believe that it is a different story. Instead of $$\binom{6}{3}\binom{3}{2}\binom{1}{1}$$ (which gives 60), we should also multiply it by 3 and then multiply it by 2, since we have to choose which task (out of 3 tasks) is assigned three people, and then we also have to choose which task (out of the remaining 2 tasks) is assigned two people. So the answer I'd like to give is 360 for this case:
$$\left[\binom{6}{3} \cdot \binom{3}{1}\right]\left[\binom{3}{2} \cdot \binom{2}{1}\right]\binom{1}{1}$$
There is also a case with 4, 1, 1 people assigned to each of three tasks, so similarly, instead of just $$\binom{6}{4}\binom{2}{1}\binom{1}{1}$$, we should also multiply it by 3 to choose which tasks (out of 3) is lucky enough to get 4 people working on it, so my answer to this case would be:
$$\binom{6}{4}\binom{2}{1}\binom{1}{1} \cdot 3$$
Answer to this problem provided by the book

Could you guys double check it with me whether my reasoning is correct?
Thanks
 A: So... rewording the question.  We have six distinct students, we have three distinct tasks.  We wish to assign students to tasks so that each task has at least one student assigned to it and each student is assigned to exactly one task a piece.
We partition the arrangements based on the sizes of the teams, and we do so specifically in decreasing order.


*

*Case $2$-$2$-$2$:  We pick the two students for the first task, two of the remaining students for the second task, and the remaining two for the final task.  $\binom{6}{2}\binom{4}{2}\binom{2}{2}=90$ arrangements.

*Case $3$-$2$-$1$:  We pick three students for the larger task and which task it is that receives the larger number of students.  We pick two of the remaining students for the medium task and we pick which remaining task receives those students.  The remaining student goes to the remaining task.  $\binom{6}{3}\cdot 3\cdot \binom{3}{2}\cdot 2=360$

*Case $4$-$1$-$1$:  We pick four students for the larger task and which task it is.  Now... there are two students left and two tasks left.  Pick which task the youngest remaining student is assigned to.  The final student will be assigned to the other task.   $\binom{6}{4}\cdot 3\cdot 2=90$
Now... if we wanted to break each of these cases down further, e.g. from $321$ into each of $321,312,123,132,213,231$... then you can divide that result by $6$... giving $60$ arrangements in each of these subcases.
If we wanted to break case $411$ down further into $411,141,114$ then we divide that result by three (not by six), giving $30$ arrangements in each of these subcases.

If we wanted to add all of these up, we get $90+360+90=540$.  If we wanted to count this directly, we could use stirling numbers of the second kind.
There are $\left\{\begin{matrix}6\\3\end{matrix}\right\}\cdot 3! = 90\cdot 6 = 540$ arrangements of $6$ distinct objects into $3$ distinct non-empty groups.
