Simple combination problem( arranging objects in rows)

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Homework Help Overview

The problem involves arranging 3 girls and 3 boys in a row with the condition that the boys must sit together. Participants are exploring the implications of this condition on the total number of arrangements.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the arrangement of boys and girls, questioning the counting of positions and the factorial calculations involved. There is a focus on understanding how to treat the group of boys as a single unit and the implications of this on the arrangement.

Discussion Status

Some participants have provided clarifications regarding the counting of arrangements, particularly about the fixed positions of the boys and the empty slots. There seems to be a productive exchange of ideas as participants refine their understanding of the problem.

Contextual Notes

There is an underlying assumption that the boys must be treated as a single unit for the arrangement, which has led to confusion regarding the counting of arrangements. Participants are working through these assumptions to arrive at a clearer understanding.

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Homework Statement

there 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?

Homework Equations

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

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

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