I need help understanding part (c) of the question below.
A group of 5 boys and 5 girls line up outside a cinema. In how many ways can they be arranged:
(a) With no restriction?
(b) If a particular girl stands in line first?
(c) If they alternate between boys and girls (with either a girl or boy in first place)?
Answers:
(a) Since there are 10 unqiue people we do 10P10 = 10! = 3,628,800
(b) _ _ _ _ _ _ _ _ _ _ ----> There are 10 spots, but the first one on the left only has 1 possibility (the one girl that has to stand first), the other spots have 9! possibilities -----> $1\times9!$ = 362,880
(c) I know how to get the answer, but I'm not quite sure why it works. So again, you have 10 spots -------> _ _ _ _ _ _ _ _ _ _ ,
Assume that we start with the girls first on the left-hand side, then there are 5 possibilities. The next has to be a boy and there are 5 boys left, so there are 5 possibilites. The next spot has to be a girl and there are 4 girls left, so there are 4 possibilities. And so on................... $\underline{5}\times\underline{5}\times\underline{4}\times\underline{4}\times\underline{3}\times\underline{3}\times\underline{2}\times\underline{2}\times\underline{1}\times\underline{1}=5!\times5!=14400$.
But you can replicate this exact scenario where the boys start first on the left-hand side and there will be the same amount of permutations. So you have $2(5!\times5!) = 28,800$ which is the correct answer. But my method of solving it this way fails when there is an unequal amount of boys and girls. For example, what if you had 5 boys and 4 girls? How would you then solve that question? Can you suggest any other methods of solving this question?
You're actually quite limited there.
If there were $6$ boys and $4$ girls, it would be impossible to alternate genders.
In the case with with $5$ boys and $4$ girls, the girls have to be lined up between the boys, which means a boy is first and last. This would leave only $5!4!$ orders.