4 men and 4 women sitting on 2 rows with 4 seats each

Question

I've come up with this problem:

4 men and 4 women are going to sit in 2 rows with 4 seats each. In how many different ways may they sit in order that in one and only one of those rows 2 men occupy the edge seats?

By my calculations it's 16 128 ways but can anyone confirm or deny this result?

My attempt:

1. I put $2$ men on the edge seats of one row and a man and a woman on the edge seats of the other row: $4 \times 3 \times 4 \times 2 \times 4!$ I have to consider the swap of woman and man so I have to multiply this by $2$.
2. I put $2$ men on the edge seats of one row and $2$ women on the edge seats of the other row $4 \times 3 \times 4 \times 3 \times 4!$

Then I add these results and multiply by $2$ to take into account the $2$ rows.

Thank you.

Answer 1

I got the same result. I listed the two rows in sequence, so MMMMWWWW represents all men in the first row and all women in the second.

Case 1: M _ _ M W _ _ W , or vice versa. In this case, for each of these positions, you have $4*4*3*3*4*2*1*3$ which results in $3,456$. But you have to multiply by $2$, since the men in the edge could be in the second row, so $6,912$.

Case 2: M _ _ M M _ _ W, or vice versa. Similarly, you have $4*4*3*3*2*2*1*4$. which results in $2,304$ But you have to multiply by $2$, since the men in the edge could be in the second row, then by $2$ again, since the man and the woman in the edge could alternate positions, so $9,216$

Adding those numbers $6,912 + 9,216 = 16,128$.

Answer 2

Your solution is correct.

Let's confirm your answer with a different approach.

Let's ignore the restriction that the men cannot occupy the ends of both rows for now. There are two ways to choose the row in which men occupy both ends, four ways to choose the man at the left end of that row, three ways to choose one of the remaining men to sit at the right end of that row, and $6!$ ways to seat the remaining six people in the six seats, which gives us a preliminary count of $$2 \cdot 4 \cdot 3 \cdot 6!$$ However, we have counted those arrangements in which the men occupy both ends of both rows twice, once for each way we would have designated one of the rows as the row in which men occupy both ends of the row. We do not want to count these arrangements at all, so we must subtract twice the number of arrangements in which the men sit at both ends of both rows.

How many such arrangements are there?

There are $4!$ ways to arrange the four men in the four seats at the ends of the two rows and $4!$ ways to arrange the four women in the remaining four seats. Hence, there $$4!4!$$ seating arrangements in which the men occupy both ends of both rows.

Therefore, the number of admissible seating arrangements is $$2 \cdot 4 \cdot 3 \cdot 6! - 2 \cdot 4!4! = 16~128$$ as you found.