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?

First, you pick the elements of each set.

You want $2$ elements of $A$. There are $^3C_2 = 3$ distinct ways of doing this. For the set $B$, you want $2$ elements. So there are $^3 C _2 = 3$ distinct ways of doing this. For the set $C$, you want $1$ element of those $2$. So there are $2$ distinct I ways of picking it. Now use the rule of the product.
So for the choice of the elements, there are $3 \cdot 3 \cdot 2 = 18$ distinct ways of picking the elements that you need

Then, you want to make an ordered sequence of these $5$ elements that you picked. Since they don’t say anything, we will assume that in fact we can mix the elements (otherwise it would be said).

This means that we have $5$ elements to put in $5$ “places” and order is relevant. So there are $5! = 120$ distinct ways placing them to construct the sequence

Finally, putting all together (and again, by the rule of prodcut), you get that there are $18 \cdot 120 = 2160$ distinct Sequences

Using the same reasoning, you pick $3$ objects from $A$ (there are only $1$ way of doing this), you pick $2$ objects from $B$ (there are $3$ of doing this) and you pick one element from $C$ (there are only one way to do this). So we get $3$ Distinct ways of pick up the elements. Then you construct the sequence, and there are $6! = 720$ Distinct ways of doing this. In total, the solution for the last part would be $3 \cdot 720 = 2160$
