Problem on fundamental counting principle.

There are 5 boys and three girls. If two girls cannot sit next to each other, how many ways can all of them sitin a row together.

I'm not really sure if what I did was right, but my answer is 240. I added 5! (number if combinations the boys can make) and 6!/(6-3)! (the number of slots between the boys distributed to the number if girls) I got 120 for each. I'm not really sure if what I did was right, or even how to check if it's right.

Re: Problem on fundamental counting principle.

Quote:

Originally Posted by

**miguel11795** There are 5 boys and three girls. If two girls cannot sit next to each other, how many ways can all of them sitin a row together.

I'm not really sure if what I did was right, but my answer is 240. I added 5! (number if combinations the boys can make) and 6!/(6-3)! (the number of slots between the boys distributed to the number if girls) I got 120 for each. I'm not really sure if what I did was right, or even how to check if it's right.

Your answer is really too low. There are two ways to approach this question.

$\displaystyle 8!-2\cdot 7!=30240$, that is subtracting the cases where they are together from the total.

Or $\displaystyle \frac{7!}{(7-2)!}\cdot 6!=30240$. The six "others" make seven places to put the two girls.