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
 A: Your answer to the first question is correct.
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.
A: 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).
A: a)  I think that answer is correct
b)The women can be arranged in 4! ways, the men can be arranged in 4! ways.
Then the men must be placed in one of the 5 following positions _ W _ W _ W _ W _   
$\therefore\,$ there are $4!*4!*5 = 2880$ ways all the men can sit together.
