Problem sets that I don't know how to answer... HELP!

Question

How many ways can 3 boys and 3 girls line up for a bus?

a.) If only the boys must be together?

b.) If no two people of the same s*x are allowed to be next to each other?

Answer 1

i say B

Answer 2

a) If the boys must be together, consider them as one 'person' temporarily. Now, you have to arrange 4 persons.

There are a total of 4! = 4*3*2*1 = 24 ways to arrange 4 persons if their order matters. (4 choices for the 1st place, 3 choices for the 2nd place, ...)

There are 3! = 6 possible arrangements of the 3 boys. (ABC, ACB, BAC, BCA, CAB, CBA)

Therefore, there are (4!)(3!) = 144 ways to arrange 3 boys and 3 girls if boys must be together.

b) You are effectively making two lines, one for boys and one for girls. e.g. B1B2B3, G1G2G3

Then put them together B1G1B2G2B3G3 or G1B1G2B2G3B3, so that no two people of the same s*x are next to each other.

There are 3! ways for the boys and 3! ways for the girls. Hence, there are a total of (2)(3!)(3!) = 72 ways to arrange them.