Four, (20 points) Question: In how many different ways can the elements of the set $M=\{1,2,3,4,5\}$ be assigned to three (ordered) sets $A$, $B$, and $C$, such that each element is contained in at least one of the sets, the intersection of these three sets is empty, and the intersection of any two of these sets is not empty?
As shown in Figure 2, consider the seven parts divided by the Venn diagram, represented by $x, u, v, w, a, b, c$ respectively.

Now, fill the elements of $M$ into these parts. According to the problem, $x$ cannot be filled with any number, while $u, v, w$ must be filled with numbers, and the numbers filled in these parts must be different (otherwise, the same elements would be placed in the $x$ region); $a, b, c$ can be filled with or without numbers, and different regions will not contain the same elements (otherwise, they would be placed in $u, v, w$).

Let $\bar{u}$ represent the number of elements filled in $u$, and similarly for others.
By symmetry, we can list the cases where $\bar{u} \leqslant \bar{v} \leqslant \bar{w}$, which results in four scenarios:
(1) $(\bar{u}, \bar{v}, \bar{w})=(1,1,1)$;
(2) $(\bar{u}, \bar{v}, \bar{w})=(1,1,2)$;
(3) $(\bar{u}, \bar{v}, \bar{w})=(1,2,2)$;
(4) $(\bar{u}, \bar{v}, \bar{w})=(1,1,3)$.
For scenario (1), take one number from $M$ and place it in each of $u, v, w$, which gives $5 \times 4 \times 3=60$ ways. The remaining two numbers can be placed in $a, b, c$ arbitrarily, which gives $3^2$ ways. Therefore, scenario (1) has $60 \times 9=540$ ways.

For scenario (2), for $u, v, w$, the grid containing two numbers has three cases. For any of these cases, take two numbers from $M$ and place them in one grid, and place one number in each of the other two grids, which gives $\mathrm{C}_{5}^{2} \mathrm{C}_{3}^{1} \mathrm{C}_{2}^{1}=60$ ways. The remaining one number can be placed in one of $a, b, c$, which gives 3 ways. Therefore, scenario (2) has $3 \times 60 \times 3=540$ ways.

For scenario (3), for $u, v, w$, the grid containing one number has three cases. For any of these cases, take one number from $M$ and place it in one grid, take two numbers and place them in one grid, and place the remaining two numbers in another grid, which gives $\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{2}=30$ ways. Therefore, scenario (3) has $3 \times 30=90$ ways.

For scenario (4), for $u, v, w$, the grid containing three numbers has three cases. For any of these cases, take three numbers from $M$ and place them in one grid; place one number in each of the other two grids, which gives $\mathrm{C}_{5}^{3} \mathrm{C}_{2}^{1}=20$ ways. Therefore, scenario (4) has $3 \times 20=60$ ways.
In summary, there are
$$
540+540+90+60=1230 \text{ (ways). }
$$