Solve Seating Arrangement of 4 Men and 4 Women No Pair Next

[Fermat1]

Suppose we have 4 men and 4 women. In how many ways can they sit in a row if no two men and no two women can sit next to each other?

The easy way to do this is simply to calculate 4! x 4! x 2.
But I was thinking about it the other way round. There are 8! possibilities for ordering 8 people so we nned to take some of those outcomes out.

The number of ways of ordering if at least 2 men (or women) must sit next to each other is I thought:

$$\frac{4C2.2.6!.7}{4.4}$$ because we can choose 2 men from 4, if we fix the 2 men the 6 others can permute how ever they wish, the 2 men can switch positions and the two men can move along the row, but we must exclude the possability that the pair is mixed sex.

This is the complement of the group sitting man-women-man-women, so the answer to the original question ought to be 8! divided by this fraction. The answer is quite a way out however. How does my reasoning err?

 

[chisigma]

Suppose we have 4 men and 4 women. In how many ways can they sit in a row if no two men and no two women can sit next to each other?

The easy way to do this is simply to calculate 4! x 4! x 2.
But I was thinking about it the other way round. There are 8! possibilities for ordering 8 people so we nned to take some of those outcomes out.

The number of ways of ordering if at least 2 men (or women) must sit next to each other is I thought:

$$\frac{4C2.2.6!.7}{4.4}$$ because we can choose 2 men from 4, if we fix the 2 men the 6 others can permute how ever they wish, the 2 men can switch positions and the two men can move along the row, but we must exclude the possability that the pair is mixed sex.

This is the complement of the group sitting man-women-man-women, so the answer to the original question ought to be 8! divided by this fraction. The answer is quite a way out however. How does my reasoning err?

The possibilities are separate groups in the following order...

a) 4 men + 4 women

b) 4 women + 4 men

c) 2 men + 4 women + 2 men

d) 2 women + 4 men + 2 women

e) 2 men + 2 women + 2 men + 2 women

d) 2 women + 2 men + 2 women + 2 men

By symmetry and possibility of reordering all these groupings have the same number of combinations so that the total number of solutions are 4! 4! 6...

Kind regards

$\chi$ $\sigma$