These problems are all sub problems of number 7, I've been assuming that they are all related, and there are no additional instructions not listed. I'm mostly just looking for confirmation that my solution is entailed from the premises.

7(b) In how many ways can 3 boys and 3 girls sit in a row if the boys and the girls are each to sit together?

I'm assuming the first slot, of six total, can be filled by anyone so there are six possibilities.

The second slot has to be filled by someone of the opposite sex so there are three possibilities.

The third and fourth slots work the same as the first two slots only there are now 4 total, and 2 of the opposite gender to select from.

Similarly for the last two slots.

So I'm claiming that this works out to be $\displaystyle (6*3)(4*2)(2*1)=288$

Since the given answer is 72 this must be wrong. And I haven't been able to work out a way that ends with the correct answer. Like I've tried a few different interpretations of the question and I still get nothing.

Also, since 288 is an integer multiple of 72, there must be a factor of 4 in my solution someplace it isn't supposed to be.

7(c) In how many ways if only the boys must sit together?

So I'd guess this works out too;

bbb,ggg

g,bbb,gg

gg,bbb,g

ggg,bbb

where any given boy or girl can be any of the three, so;

(3*2*1)(3!)

3*(3!)*(2!)

3*2*(3!)*1

3!*3!

or just $\displaystyle (3!*3!)*4=144$ which is listed as the correct answer. Is this solution correct?

7(d) In how many ways if no two people of the same sex are allowed to sit together?

This one seems strange; is boy, boy, boy a good configuration? As no two people of the same sex are sitting together.

b,g,b,g,b,g which is $\displaystyle 3*3*2*2*1*1=36$

and alternately

g,b,g,b,g,b

For a total of 72 combinations, which is the listed answer. Is this solution correct?