Permutations Problem

Hi, i'm having a bit of an issue with solving a permutations problem

Find the number of ways in which 4 boys and 4 girls can be seated in a row of 8 seats if they sit alternately.
Okay, well.. Simple enough, after solving the permutations of both cases I get 1152 permutations

There is a boy named Micah and a girl named Loretta in problem number 4, and they cannot be seated next to each other or else they will fight. How many permutations are there now?

I'm completely stuck on this one, without any idea on how to start except for getting the number of permutations where they *are* sitting together.

Any help would be appreciated (not just an answer please!)

agree on the 1152. 8•4•3!•3! = 32 • 36 = 1152

for the 2nd, let's count the number of ways Micah and Loretta can sit together and then subtract them out. 2 ways they can be in positions 1&2, and then 3! • 3! ways to fill the other spots, 72 ways.
same for positions 2&3, 3&4, ... 7&8, so a total of 7 • 72 = 504. subtracting leaves 648 ways