Simple combination problem( arranging objects in rows)

  • Thread starter torquerotates
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    Combination
In summary, there are 3 girls and 3 boys who need to sit in a row. If the boys must sit together, there are (3!)(4!) ways for the boys and girls to be arranged. This takes into account the 3! ways of arranging the boys within each group of 3 and the 4! ways of arranging the groups of boys within the row. The 3! ways of arranging the girls is also taken into consideration.
  • #1
torquerotates
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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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  • #2
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
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
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
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
That's it.
 

1. How many ways can objects be arranged in a single row?

The number of ways that objects can be arranged in a single row is equal to the factorial of the number of objects. For example, if there are 4 objects, there are 4! = 4 x 3 x 2 x 1 = 24 ways to arrange them in a row.

2. What is the difference between a permutation and a combination?

A permutation is an arrangement of objects in a specific order, while a combination is a selection of objects without regard to order. For example, a permutation of ABC could be ABC, BAC, or CBA. A combination of ABC could be ABC, ACB, BAC, BCA, CAB, or CBA.

3. How does the number of objects affect the number of possible combinations?

The number of objects affects the number of possible combinations by increasing the number of options for each position in the combination. For example, if there are 3 objects to choose from, there are 3 options for the first position, 2 options for the second position, and 1 option for the third position, resulting in 3 x 2 x 1 = 6 possible combinations.

4. Can objects be repeated in a combination?

Yes, objects can be repeated in a combination. This is known as a combination with repetition or a combination with replacement. For example, a combination of 3 letters from the alphabet with repetition would include AAA, AAB, AAC, and so on.

5. How can I calculate the number of combinations if I have more than one type of object?

If you have more than one type of object, you can calculate the number of combinations by finding the number of combinations for each type and then multiplying them together. For example, if you have 3 red objects and 2 blue objects, there are 3 ways to choose a red object and 2 ways to choose a blue object, resulting in 3 x 2 = 6 possible combinations.

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