# Arranging problem: 4 couples, 8 seats in a row… Am I making this too simple?

I am in a prob and stats course... haven't taken one in awhile and would like some help on these two problems. I think I am probably making these a little two simple.

Four married couples have bought 8 seats in a row for a concert. In how many ways can they be seated if:

a. if each couple sit together? 4!2!2!2!2!= 384

b. if all men sit together?

I am thinking of this M1M2M3M4W1(Any women)(Any Women)(Any Women)(Any Women) so 4*3*2*1*4*3*2*1= 576

• One way to check your idea is to use the same methods to solve a simpler problem where the answer will be clear. Do your methods give the right answers when there is only one or two couples? – MJD Jun 9 '15 at 2:12

Your first answer is correct: there are $4!$ ways to order the couples, treating each as a unit, then you can flip the order of each member of a couple, so the answer is multiplied by $(2!)^4$.

For your second answer, I believe you are undercounting. You also need to consider configurations like $M_1(...\text{women}...)M_2M_3M_4$.

Think about it like this: first the men sit down in a block

oooo


there are $4!$ ways to do this. Then the women choose a space between two men: there are $5$ spaces to choose from.

xoooo
oxooo
ooxoo
oooxo
oooox


Now order the women: there are $4!$ ways to do this.

So the answer is $5 \cdot 4! \cdot 4!$, or $5!\cdot4!$ (but I would prefer the first version since it emphasizes where the $5$ comes from).

• One observation, all the women sit together is what you did. He asked for the men, which is mathematically no different. – FundThmCalculus Jun 9 '15 at 2:55
• You are correct, excuse me! – Eli Rose Jun 9 '15 at 2:56
• No problem, just thought I would make that known for the benefit of the original poster. Excellent answer with nice graphical depications. – FundThmCalculus Jun 9 '15 at 10:16

For the second question, think of the four men as a unit. That gives you five objects to arrange, the block of four men and the four women. The five objects can be arranged in $5!$ ways. Within the block of four men, the men can be arranged in $4!$ ways. Therefore, there are $4!5!$ seating arrangements in which the men sit together.
• Ah, thinking of it as the placement of $5$ objects is quite nice. – Eli Rose Jun 9 '15 at 2:19
$\therefore\,$ there are $4!*4!*5 = 2880$ ways all the men can sit together.