# Combinatorics: “four male-female couples and eight chairs” problem

I was trying to solve the following problem:

Eight persons, consisting of four male-female couples, are to be seated in a row of eight chairs. How many seating arrangements are there in each of the following cases:

1. There are no other restrictions.
2. The men must sit together and the women must sit together.
3. The men must sit together.
4. Each couple must sit together.

For the first and forth points I was able to find a solution easily (respectively $10*9*8$ and $8*6*4*2$), but for the other middle points I am having some little troubles. Doesn't the 3rd point imply the second?

In both cases, the first person will have $8$ places to choose from, but the second person that seats will have to be careful depending on the first person was a male or a female and depending if it is a male or a female.

1. $8!$, on first chair can seat 8 different people, on 2nd 7, $\dots$
2. There are two options: $MMMMFFFF$ or $FFFFMMMM$. Either way there are $4!$ possibilities to arrange males, the same goes for females. Total $2\cdot 4! \cdot 4!$.
3. Similarily as in 2., but there are 5 ways to put all the man together, and $4!$ ways to arrange them in a group (same for females), so $5 \cdot 4! \cdot 4!$.
4. $4! * 2^4$, $4$ couples, each couple takes 2 chairs, 2 ways to arrange each couple.
• I don't understand your third point. Why do we have $5$ ways to put the men together? – nbro Jan 2 '16 at 13:52
• I have another related question. How would you proceed with your reasoning in a general case where we have $n$ chairs $f$ females and $m$ males for the 3rd case? – nbro Jan 2 '16 at 14:37