Having an issue with a permutation question.
$3$ boys $4$ girls need to sit down in a row
(a) How many ways can you arrange them if there are no restrictions?
$7! = 5040$. I think anyway!
(b) How many ways can you arrange them if the girls have to sit together?
$(4 \times 3 \times 2 \times 1) \times3 \times2 \times 1 = 144$ permutations. Not sure on this one. Do i count the boys as a seperate entity and only allow one permutation?
(c) How many ways can you arrange them if no two girls can sit beside each other?
My thoughts on this are $4\times3\times3\times2\times2\times1\times1$
I have put the numbers that i think represent the girls in italics
Thanks for any help!
I'm assuming by your answers that you are arranging these people side-by-side instead of circularly, and that you distinguish between different boys and girls.
How many ways can you arrange them with no restrictions?
You're correct, there are $7!$ ways to arrange them.
How many ways can you arrange them if the girls have to sit next to each other?
Then we can treat all the girls as though they are one entity, so there are $4!$ ways - three boys, and one "girl". The "girl", however, can be arranged in $4!$ ways. Thus there are $4! * 4! = 2 * 4!$ ways to arrange them.
How many ways can you arrange them if no two girls can sit beside each other?
There are four girls and three boys, so necessarily the arrangement looks like:
$$YXYXYXY$$
Where $X$ is a boy and $Y$ is a girl. There are $3!$ ways to arrange the boys, and $4!$ ways to arrange the girls, so the total number of arrangements is $3! * 4!$.