(11. Determine how many different ways there are to assign the elements of the set $M=\{1,2,3,4,5\}$ to three (ordered) sets $A, B, C$, such that each element is contained in at least one of the sets, the intersection of the three sets is empty, and the intersection of any two sets is not empty? (i.e., $A \cup B \cup C=M, A \cap B \cap C=\varnothing$, and $A \cap B \neq \varnothing, B \cap C \neq \varnothing, C \cap A \neq \varnothing$)
11 Consider the seven parts divided by the Venn diagram, represented by $x, u, v, w, a, b, c$ respectively. Now, elements of $M$ are to be filled 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 elements filled must be distinct (otherwise, identical elements would be placed in the $x$ region); $a, b, c$ can be filled or left empty, but they cannot 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 the others. By symmetry, we can list the cases with $\bar{u} \leqslant \bar{v} \leqslant \bar{w}$, which results in four cases:
(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 (1), one number is taken from $M$ and placed in each of $u, v, w$, which can be done in $5 \cdot 4 \cdot 3=60$ ways. The remaining two numbers can be placed in $a, b, c$ arbitrarily, which can be done in $3^2$ ways. Therefore, the number of ways corresponding to (1) is $60 \times 9=540$.

For (2), there are 3 cases where two numbers are placed in one of $u, v, w$. For any of these cases, two numbers are taken from $M$ and placed in one part, and one number is placed in each of the other two parts, which can be done in $\mathrm{C}_{5}^{2} \mathrm{C}_{3}^{1} \mathrm{C}_{2}^{1}=60$ ways. The remaining one number can be placed in one of $a, b, c$, which can be done in 3 ways. Therefore, the number of ways corresponding to (2) is $3 \times 60 \times 3=540$.

For (3), there are 3 cases where one number is placed in one of $u, v, w$. For any of these cases, one number is taken from $M$ and placed in one part, two numbers are taken and placed in another part, and the remaining two numbers are placed in the last part, which can be done in $\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{2}=30$ ways. Therefore, the total number of ways is $3 \times 30=90$.

For (4), there are 3 cases where three numbers are placed in one of $u, v, w$. For any of these cases, three numbers are taken from $M$ and placed in one part, and one number is placed in each of the other two parts, which can be done in $\mathrm{C}_{5}^{3} \mathrm{C}_{2}^{1}=20$ ways. Therefore, the total number of ways is $3 \times 20=60$.

Combining all, the total number of ways is $540+540+90+60=1230$. Each of these ways results in three sets $A, B, C$ that satisfy the conditions. Therefore, the answer to this problem is 1230.