# Arranging men and women in rectangular table, so that in front of each woman is a man.

A group of $$4$$ men and $$4$$ women need to be distributed on two opposite sides of a rectangular table, so that in front of each woman is a man. How many options are there?

My attempt:

I thought that if a man must be in front of a woman, I could assume that the number of combinations is $$4 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 1 \cdot 1$$ or $$4!4!$$, since if I pick a side and distribute them among the other sides, there will be one person less for each side.

Although, I'm not sure if it's right because it can happen two women to be seated side by side.

I would appreciate any help that clarify me the problem.

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Let's say that the positions are numbered like this:

$$1|2|3|4$$

$$5|6|7|8$$

That means that the seat $$5$$ is opposite of seat $$1$$, seat $$6$$ is opposite of seat $$2$$, etc.

You can chose between $$8$$ people for whom to put in seat number $$1$$.

After this, you can chose between $$4$$ people for whom to put opposite of seat number $$1$$(that is in seat number $$5$$).

Next you can chose between $$6$$ people for whom to put in seat number $$2$$, and then you can chose between $$3$$ people to put in seat number $$6$$

Continuing this way we can find that there are $$8*4*6*3*4*2*2*1$$ ways to arrange the people the way you want.

There are $$4!$$ possible arrangements for $$4$$ (M,W)-pairs.
Placing the women one by one there are $$8$$ spots for the first, $$6$$ for the second, $$4$$ for the third and $$2$$ for the last.
So we arrive at:$$4!\times8\times6\times4\times2=9216$$ possible arrangements.