permutation & combinations

[opponent]

In how many ways can 4 men and 4 women sit at a round table with no two women in consecutive positions?

[maihuna]

Let W, M means women men
W/o no constraint 8 people on a round table can seat in !7 ways

But with no two women together constraints:
wxwxwxwx in 4*4*3*3*2*2 = 576

since entries are round we do not need to multiply by 2 ways

[Night reader]

In how many ways can 4 men and 4 women sit at a round table with no two women in consecutive positions?

MAN-A, WOMAN-B, A-B-A-B-A-B-A-B OR B-A-B-A-B-A-B-A. Since we start with A and end with B in a circular permutation where the end of one order is the beginning of the other, either MAN or WOMAN element must decrease by One 4!*3!=144

[Night reader]

Hi Maihuna, if that would be permutations in a row then yes the set wxwxwxwx can be different from xwxwxwxw or if you put numbers w1x1w2x2w3x3w4x4 is different from w4x4w1x1w2x2w3x3 or even w3x3w4x4w1x1w1x1 BUT in a circular order (permutations) the listed sets would be considered as the same unless there are relative changes within the sets i.e. - w1x2w2x3w3x4w4x1. So you have over-counted by exactly 4 times ways or 576/4=144

Let W, M means women men
W/o no constraint 8 people on a round table can seat in !7 ways

But with no two women together constraints:
wxwxwxwx in 4*4*3*3*2*2 = 576

since entries are round we do not need to multiply by 2 ways