There are four couples to be seated in a row with $8$ seats.

Question

There are four couples to be seated in a row with $8$ seats.

the number of ways they can seat

(a) without any restriction is $8!$.

(b) each couple is seated together is $4!$.

(c) the males and females have to seat together is $(4! \times 4!) \times 2$.

Is the solution correct?

Answer 1

Your answer for b is a bit off.

If you treat each couple as a block, then you can arrange the blocks in $4!$ ways.

However, you can arrange each couple INSIDE the block in $2$ ways.

There are $4! \cdot 2^4$ ways to have each couple sitting together.

Answer 2

your part $b$ is not right

couples can also permute between themselves in $2!$ ways so better multiply your answer in part$\ b\ $by $(2!)^4$ rest is fine