Basic counting trouble Suppose you have $8$ people. How many possible ways can you seat these people if 
a) two persons, A and B, must sit together? 
b) $4$ men and $4$ women, but no $2$ men and no $2$ women can sit together  
c) there are $5$ men who must all sit together
d) there are $4$ married couples that must sit together
I suppose I'm confused as to what my error is on some of these by doing:
a) This makes it so that we are dividing the seating into a group of $2$ and $6$ groups of $1$, so we have $$\frac{8!}{2!}$$ ways to do this. But also we could view this as only having $7$ degrees of freedom, twice, so we'd have $$2 \cdot 7!$$. Which one is correct and why is the other wrong??
b) This one I do not know how to do.
c) Again, is this not just a group of five and three groups of 1? to get
$$\frac{8!}{5!}?$$ Or is it $4! \cdot 5!$ since we have 4 degrees of freedom and $5!$ distinguishable items?
d) Again, is this
$$\frac{8!}{2^4}$$ or is it $4$ groups of $2$ giving $4!2^4$?? Why do both answers here seem right??

Are these circular tables or tables with a distinct orientation.
Based on the way you answered the first question, I am going to guess that these tables have an orientation.
a) 1 group of 2 and 6 groups of 1 is 7 groups total.
$2\cdot 7!$
b) You can but a man at the head or you can put a woman a the head.  Once you have made this decision there are $4!$ to arrange the men and $4!$ ways to arrange the women.
c) there is a group of 5 and 3 groups of 1.  $5!\cdot 4!$
d) 4 groups of 2. $(2!)^44!$
If you descide that this is a circular table.  Then for any of the configurations above, you can always rotate the table such that Mr. Jones is at the head.
Divide the results by $8.$
