Arrangement in a row In how many ways can $8$ people be seated in a row if
(a) There are $4$ men and $4$ women and no $2$ men or $2$ women can sit next to each other?
(b) There are $5$ men and they must sit next to each other?

I got $4!\cdot{5\choose 4}\cdot4!$ for first one and $5!\cdot4!$ for second one.
I am a bit sceptical about it .
 A: 
$(a)$ Consider the first chair. We can put there a man or a woman.
Suppose we put one of the $4$ men. Then for the second chair, we must put one of the $4$ women. For the third chair, we will have to put one of the $3$ remaining men. And so on, until we fill the row.
By the product rule, there are $4^2 \cdot 3^2 \cdot 2^2 \cdot 1^2 = 576$ distinct ways of doing this.
The same argument applies if we put a woman in the first chair.
So the total ways of sit these $8$ people in this condition is $2 \cdot 576 = 1152$.



$(b)$ We have $8$ chairs and we want to put $5$ men together. So in terms of their position, we can put such a group in $4$ distinct ways (starting in first chair until the fifth chair, starting in the second chair until the sixth chair, and so on).
These men are able to switch places between them, so the group of $5$ men can sit in their $5$ chairs in $5!=120$ distinct ways.
Next, we have to put the remaining $3$ people in $3$ chairs and they can switch places between them, so there are $3! = 6$ distinct ways of placing them.
By the product rule, we conclude that are $4 \cdot 5! \cdot 3! = 4 \cdot 120 \cdot 6 = 2880$ distinct sequences in this condition.


In your answer, part $(a)$ is wrong and part $(b)$ is right.
