5 seats and 2 girls have to be together how many ways can the group of two girls be arranged (not counting the fact that the two girls are different)
this gives 4 possibilities
xx000 0xx00 00xx0 000xx (where x's are girls 0's are guys)
But there are two different girls so we have twice as many ways to arrange the girls!
so 8 possibilities
now the boys are all different, how many ways can we arrange the three boys (6 ways or 3!)
I only get this part of what you're saying (other part might be right but I dont get it). Basically what I'm saying is that you get those 4 arrangements (demonstrated by the x's and o's) and then accounting for the fact that your x's are 2 different girls you get two possibilities for each of the 4 arrangements (resulting in 8). Now you do the same thing for the boys (you have 3 o's but 3 diff boys!) and using the same logic you get 6 different ways for each of the 8 arrangements. this results in 6*8 =48 ways.