I designed a question where there are 12 employees, and they are to be assigned to 3 distinct teams (let's call them team 1, 2 and 3). Suppose the twelve people are $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$, $X$, $X$, and $X$ (for brevity I'm not adding the whole question here but this is its basic design).
I wanted to know how many different teams can be formed such that all the $X$'s are in one team. Then I wanted to find the probability that all $X$'s go into one team. I started out grouping the $X$'s in one team at a time, and using the concept of partitioning the employees into teams. I'll summarize my approach below:
Let's define the events $E_{X,n}=\text{\{All X's go to team n\}},n=1,2,3$.
$$\text{Total number of different teams of those employees} = \binom{12}{4,4,4} = |\Omega|$$
$$\text{Number of different arrangements of all employees in teams such that all X's go to team 1}=1\times \binom{9}{1,4,4} = |E_{X,1}|$$
$$\text{Probability that all X's go into team 1} = \frac{|E_{X,1}|}{|\Omega|}$$
$$\text{Number of different arrangements of all employees in teams such that all X's go to team 2}=1\times \binom{9}{1,4,4}=|E_{X,2}|$$
$$\text{Probability that all X's go into team 2} = \frac{|E_{X,2}|}{|\Omega|}$$
$$\text{Number of different arrangements of all employees in teams such that all X's go to team 3}=1\times \binom{9}{1,4,4} = |E_{X,3}|$$
$$\text{Probability that all X's go into team 3} = \frac{|E_{X,3}|}{|\Omega|}$$
since there is only one way to combine all the $X$'s into any team, and the rest can join in any order.
I can't figure out of all the $E_{X,n}$ are disjoint or not, because I don't know what the sample space will be like.
As you solved, the probability $P(A) = \displaystyle \frac{3 \times \binom{9}{1,4,4}}{\binom{12}{4,4,4}} = \frac{3}{55}$.
The other way to look at it is that whichever team first $X$ gets assigned to, there are $3$ more places on that team. The probability that remaining two $X$ will get $2$ of those $3$ seats $= \displaystyle \frac{3}{11} \times \frac{2}{10} = \frac{3}{55}$.