# Simple combination problem( arranging objects in rows)

1. Sep 8, 2008

### torquerotates

1. The problem statement, all variables and given/known datathere are 3 girls and 3 boys that are to sit in a row. How many ways are there for them to sit in a row if the boys must sit together?

2. Relevant equations n!=n(n-1)(n-2).....(3)(2)(1)

3. The attempt at a solution

there are 4 possible positions for the boys. o 1 o 1 o 1 o

the o's represent the group of 3 boys and the 1's each represent a girl. Within a o, there are 3! ways to arrange boys. And since there are 4 o's, there are 4! ways of arranging the o's. Just looking at the o's, we have, by counting, (3!)(4!) ways of arranging the boys. And since there are 3 1's, there are 3! ways of arranging the girls. The solution by counting is that there are (3!)(4!)(3!) ways for the boys and girls to be arranged in a row.

But the book has it at (3!)(4!). Did I do something wrong?

2. Sep 8, 2008

### Dick

Only one of your o's contains any boys in any given seating arrangement. Why are you counting different arrangements of the o's??

3. Sep 8, 2008

### torquerotates

because first I fix a o. Within that o, there are 3! ways to rearrange the boys. There are also 4! ways rearranging the o within the fixed 3 girls. But there are also 3! ways of rearranging the girls.

4. Sep 8, 2008

### Dick

You said there were 4 positions the boys could occupy. Not 4!. Three of the o's are empty in any given arrangement. Not much point in rearranging those.

5. Sep 8, 2008

### torquerotates

oh, I get it( stupid me). So 3! ways of rearranging the boys in each o and 4 ways of rearranging the o's and 3! ways of rearranging the girls give rise to 3!4!?

6. Sep 8, 2008

That's it.