arrangement of NOT sitting together I have the following exercise. Please help me to solve it.
Exercise. In how many ways can 3 men and 3 women be seated at a round table if 
(a) no restriction is imposed
(b) 2 particular women must not sit together
(c) each woman is to be between 2 men.
Ideas
(a) simple case, it's just $5!$, I am still hardly getting the idea of round table and the first man taking any place. We don't count him because the table is round is it correct? So the table doesn't have the beginning and end, so we can start comparing permutation from any of the guests, therefore we don't count the first man, I still don't have good understanding why it happens.
(b)
I've never saw the template for "must not sit together", usually when  the is a group that must sit together we take them as one guest and on addition count the permutation within the group, but here I don't know to reason about the solution. 
(c) extremely hard, I even don't have ideas.
I would appreciate for any help.
 A: Problem 1
You guessed right the answer is $5!$. The reason why we don't count the first is because the table doesn't have a beginning nor end. So let's say one combination is:
$${1,2,3,4,5,6}$$
Where men are numbered from $1-3$ and women are number from $4-6$. If we start counting from the last one we'll get a combination:
$${6,1,2,3,4,5}$$
But these combination are completely the same, so to avoid double counting we fix one as our stating point.
To make it even simpler. There are $6!$ combinations, but note there are 6 people and we can start counting from anyone so every combination will be included six times, so we divide by $6$. That's:
$$\frac{6!}{6} = 5!$$
Problem 2
It'll be much easier to count the "bad" combinations and subtract them from the total number of combinations. We could think of them two as one person, so there will be 5 people to sit down.
This means that there are $5!$ combination. But for the same reason we mentioned earlier we'll divide that number by 5. But note that when we "split" the two women the can sit in two different ways. So we'll have to multiply the number of combinations by 2. So the number of "bad" combinations is:
$$\frac{5!}{5} \cdot 2 = 4!\times2$$
And the total number of combinations that satisfy the condition is:
$$\frac{6!}{6} - 4!\times2 = 120 - 48 = 72$$
Problem 3
This is a type of stars and bars combinatorics, but because there are 3 men and 3 women, that implies that the order must be alternating, so it'll be easier to come up with our own "formula"
Let $1,3,5$ be "male places" and $2,4,6$ be "female places". For the male places there are $3!$ number of ways, so does for the female places. But note that this time we have to divide by $3$, as there are 3 persons to start counting from. So the total number of combination is:
$$\frac{3!\cdot3!}{3} = 2! \cdot3! = 12$$
