Heey how would you do these:
in how many ways can 6 boys and 2 girls be arranged in a row if the two boys:
a)are not together?
b)there are at least 3 girls between the boys
Sorry, the question doesn't make sense. I assume you mixed up the boys/girls? So, rewriting,
In how many ways can 6 girls and 2 boys be arranged if:
a) The boys are not together
b) There are at least 3 girls between the boysLet's start with a)
Without worrying about the boys being together part, there are
ways of arranging the people could be arranged. Then, we need to subtract the number of times that the boys are together.
Let's pretend that the boys ARE together. Then, we can essentially think of them as 'one entity': ie. there are 6 girls, and 1 'boy' entity. So, there are 7! ways of arranging this, multiplied by the two ways the boys could be arranged.
So, the answer will be the total number of ways, minus the number of ways that the boys ARE together. Thus,
Now, let's look at part b)
There are three possible cases; three girls are between the boys, four girls are between the boys, five girls are between the boys and six girls are between the boys. This is a bit unmanageable; it is easier to look at how many ways there could be two girls between the boys, one girl between the boys, and no girls between the boys! Then, we can subtract this from the total 8! ways we could arrange the people.
I have to run, but try thinking about these subcases, and how you would go about calculating them