Counting problem assigning student to groups Say I want to assign $4$ students $a,b,c,d$, to three teams, and each team needs at least one student in it. Only one issue, $a$ and $b$ cannot be in the same team. How many ways are there to assign teams these teams?
The way I went about this problem is lets say we have the set of students $\{a,c,d \}$ and we assign these students to a group using the permutation $P(3,3)$, then we can assign $b$ to a team, $P(1,1)$. Together we get $6 \cdot 1$ and there is $2!$ ways to do this because we do the first set mentioned or take another set $\{b,c,d \}$ and do the same thing we did above.
All together we get $36$ ways of arranging these teams.
Is this a correct way of going about this question?

Select $2$ out of $4$ students $\{a \ b \ c \ d \}$ who will be together in a group. That is $\displaystyle {4 \choose 2} = 6$ ways. One of these selections will have $a$ and $b$ together. Taking that out, we have $5$ acceptable ways to make three groups with $2$ students in one and remaining $2$ automatically in two individual groups.
Now if they are being assigned to distinct teams, we can permute the groups in $3!$ ways.
So the answer is $5 \cdot 3! = 30$ ways.
