In how many ways can four married couples attending a concert be seated in a row of eight seats if

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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

there are $4!$ ways to do this. Then the women choose a space between two men: there are $5$ spaces to choose from.

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).

You can put this solution on YOUR website!

Total number of people = 8 a) 8! b) 2* 2 * 2 * 2 * 4! c) 4! * 4! * 2!

if any doubt, feel free to ask..