11. (25 points) For a set $M=\left\{p_{1}, p_{2}, \cdots, p_{2_{n}}\right\}$ consisting of $2 n$ prime numbers, its elements can be paired to form $n$ products, resulting in an $n$-element set. If
$$
\begin{aligned}
A & =\left\{a_{1} a_{2}, a_{3} a_{4}, \cdots, a_{2 n-1} a_{2 n}\right\} \\
\text { and } \quad B & =\left\{b_{1} b_{2}, b_{3} b_{4}, \cdots, b_{2 n-1} b_{2 n}\right\}
\end{aligned}
$$

are two $n$-element sets obtained in this way, where
$$
\left\{a_{1}, a_{2}, \cdots, a_{2 n}\right\}=\left\{b_{1}, b_{2}, \cdots, b_{2 n}\right\}=M,
$$

and $A \cap B=\varnothing$, then the set pair $\{A, B\}$ is called a "couplet" formed by $M$ (for example, from the four-element set $\{a, b, c, d\}$, three couplets can be formed:
$$
\begin{array}{l}
\{a b, c d\} \sim\{a c, b d\}, \\
\{a b, c d\} \sim\{a d, b c\}, \\
\{a c, b d\} \sim\{a d, b c\} .
\end{array}
$$

Find the number of couplets that can be formed from the six-element prime set $M=\{a, b, c, d, e, f\}$.
11. Six elements can form fifteen different "slips of paper," listed as follows:
$$
\begin{array}{l}
\{a b, c d, e f\},\{a b, c e, d f\},\{a b, c f, d e\}, \\
\{a c, b d, e f\},\{a c, b e, d f\},\{a c, b f, d e\}, \\
\{a d, b c, e f\},\{a d, b e, c f\},\{a d, b f, c e\}, \\
\{a e, b c, d f\},\{a e, b d, c f\},\{a e, b f, c d\}, \\
\{a f, b c, d e\},\{a f, b d, c e\},\{a f, b e, c d\} .
\end{array}
$$

Consider the slip of paper at the intersection of the $i$-th row and the $j$-th column as a coordinate point, denoted as $(i, j)$.

For the point $(1,1)$ in the first row, it pairs with two points in each of the following rows:
$$
\begin{array}{l}
(1,1) \sim(2,2),(1,1) \sim(2,3) ; \\
(1,1) \sim(3,2),(1,1) \sim(3,3) ; \\
(1,1) \sim(4,1),(1,1) \sim(4,2) ; \\
(1,1) \sim(5,1),(1,1) \sim(5,2) .
\end{array}
$$

This results in $4 \times 2=8$ pairs.
Similarly, the points $(1,2)$ and $(1,3)$ each pair with 8 points in the following four rows.

Thus, the three points in the first row form $3 \times 4 \times 2=24$ pairs with the points in the following four rows;

Each point in the second row pairs with two points in each of the following rows, resulting in $3 \times 3 \times 2=18$ pairs;

Each point in the third row pairs with two points in each of the following rows, resulting in $3 \times 2 \times 2=12$ pairs;

Each point in the fourth row pairs with two points in the following row, resulting in $3 \times 1 \times 2=6$ pairs.

Therefore, the total number of pairs is $6 \times(1+2+3+4)=60$, meaning that 60 couplets can be created from the set $M$.