Probability? Permutations?

Question

How would you use permutations to find the arrangements possible if a line has 4 girls and 4 boys

a) alternating boy girl boy girl?

b) 2 girls want to stand together?

c) all the boys stand together?

d) probability that 3 particular people will stand together if the queue is formed randomly

Answer 1

a) Positions 1,3,5,7 are boys and
positions 2,4,6,8 are girls. Each can be arranged in 4! (24) different ways, so...

24*24 = 576

If you can alternate starting with a girl rather than a boy, solution is 576*2 = 1152

b) 2 specific girls want to stand together? In that case we consider them as 1 unit, leaving us with 7!, but...
for each case the 2 girls can be arranged in 2! different ways, so 7!*2! = 10,080

c) Similar to b) only counting the 4 boys as 1 unit leaves us with 5!, and 4! different ways to arrange the boys, so...

5!*4! = 2680

d) There are 6 different placements for the 3:

1,2,3 - 2,3,4 - 3,4,5 - etc.

The possible combinations for 1 of these is 3!*5!
so 3!*5!*6/8!

Answer 2

the general rule for permutations is n!/[(n-r)!] If two or more in a group stand together, you treat them as one in which combinations are allowed withing the group. For example, if all the boys are together, then they can all switch places as one with the girls, but you'll have to designate a separate equation to allow them to rearrange within their group.