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now consider one position between each two adjecent men. 4 women need to sit in these 4 spots in order for condition to be valid (no two men together). this can, again, be done in:
4P4 = 24 ways.

in order to have 4 women and 4 men together at a table without adjecent men, there is 1 general layout but in reality 2 configuration : MWMWMWMW or WMWMWMWM

4 positions for men and they are 4 so : 4! ways (=24)
4 positions for women and they are 4 so : 4! ways (=24)

Take this problem as an example: Arrange 3 men and 3 women to a round table so that no two men are adjacent.

There are 2*3! way to arrange the men to the table (to 1st, 3rd and 5th AND to 2nd, 4th and 6th) (let's call these 3 men A, B and C)
After arranging the men, there are 3! way to arrange the women to the table (call them 1, 2 and 3)

However, this is a round table, in which the arrangement A1B2C3 happens 6 times in 2*3!*3! of arrangements. Similar to other arrangement.

Therefore, the answer for my example is 2*3!*3!/6 = 3!*3!/3 = 3!*(3-1)!

Similarly, the answer for the posted question is 4!*(4-1)!

However, this is a round table, in which the arrangement A1B2C3 happens 6 times in 2*3!*3! of arrangements. Similar to other arrangement.

Therefore, the answer for my example is 2*3!*3!/6 = 3!*3!/3 = 3!*(3-1)!

Sorry, I still try to get this one
We are talking about arragement , right ? It should be permutations no ? I don't understand why you need to divide by 6 ?
For me the total number of possibilities is 2*3!*3!
If someone can explain please

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