14. 4 couples go to the movies, 8 people sit in a row. If each woman's neighbor can only be her husband or another woman, how many seating arrangements are there?
14. First arrange the women, there are 4! ways.

If a man sits between women (including his wife), there must be at least two men. Similarly, if a woman sits between men, there must be at least two women.

If we consider a group of women sitting together as one unit, then the 4 women can be grouped in 5 ways: $4, 3+1, 2+2, 2+1+1, 1+1+1+1$. A woman sitting alone must be at one of the ends of the row. Therefore, the grouping $1+1+1+1$ is not feasible;
When women are grouped as $2+1+1$, the women must be at the ends, and the men can only be grouped as $2+2$, the arrangement is “woman-man-woman-woman-man-woman-woman”, and there is only 1 way to arrange the men;

When women are grouped as $2+2$, there are seven arrangements: “woman-woman-man-man-man-man-woman-woman”, “woman-woman-man-man-man-woman-woman-man” or “man-woman-woman-man-man-man-woman-woman”, “woman-woman-man-man-woman-woman-man-man” or “man-man-woman-woman-man-man-woman-woman”, “man-woman-woman-man-woman-woman-man”, the men can be arranged in $2, 1, 1, 1, 1, 1$ ways;

When women are grouped as $3+1$, there are six arrangements: “woman-woman-woman-man-man-man-man-woman” or “woman-man-man-man-man-woman-woman-woman”, “man-man-woman-woman-woman-man-man-woman” or “woman-man-man-man-woman-woman-woman”, “man-woman-woman-woman-man-man-man-woman” or “woman-man-man-man-woman-woman-woman”, the men can be arranged in $2, 2, 1, 1, 1, 1$ ways;

When all 4 women sit together, there are five arrangements: “woman-woman-woman-woman-man-man-man-man” or “man-man-man-man-woman-woman-woman-woman”, “man-man-man-woman-woman-woman-woman-man” or “man-woman-woman-woman-woman-man-man-man”, “man-man-woman-woman-woman-woman-man-man”, the men can be arranged in $3!, 3!, 2!, 2!, 2!$ ways.
Thus, the total number of arrangements is $4!(1+2+5 \times 1+2 \times 2+4 \times 1+2 \times 3!+3 \times 2!)=24 \times 34=816$.