Simple permutations: approach and proof explanation for an exercise not very clear Set $A = (a_1, a_2, a_3), B = (b_1, b_2, b_3)$ and $C = (c_1, c_2)$ are assigned.
How many are the ordered sequences formed by $5$ distinct elements containing $2$ elements of $A$, two elements of $B$, and one of $C$?
Related question: Is there use of the numbers of Stirling and of Bell for a combinatorial exercise? for to help a student.

Surely there are simple dispositions of $3$ class $2$ elements without repetition that allow me to exchange the elements of each set
$$\text{For the set $A$:}\quad D_{3,2}(A)=\frac{n!}{(n-k)!}=\frac{3!}{1!}=6$$
$$\text{For the set $B$:}\quad D_{3,2}(B)=\frac{n!}{(n-k)!}=\frac{3!}{1!}=6$$
$$\text{For the set $C$:}\quad D_{2,1}(C)=\frac{n!}{(n-k)!}=\frac{2!}{1!}=2$$
Hence, if $N$ is the total number of dispositions then I will have $N=D_{3,2}(A)\cdot D_{3,2}(A)\cdot D_{2,1}(C)=72$.
From the text of the exercise I do not know if I can mix the elements of each set. What would be the solutions done from six elements built by three elements of $A$, two elements of $B$ and one of set $C$. And if I can mix them, what happen?

Your computation assumes that the five element sequence has the $a$'s first, the $b$'s second, and the $c$ last.  If you are allowed to mix them up, it is easier to choose the two $a$'s without order in ${3 \choose 2}=3$ ways, the two $b$'s without order in ${3 \choose 2}=3$ ways and the $c$ in ${2 \choose 1}=2$ ways for a total of $18$ collections of elements.  They can be placed in order in $5!=120$ ways, so there are $18 \cdot 120=2160$ five element sequences.
