Question on permutation and combination(girls and boys standing in a line)

[mutineer123]

Homework Statement

http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w03_qp_6.pdf
question 6 b ii

Homework Equations

The Attempt at a Solution

The concept is bothering me. the answer is 5! × 3! × 4C1. Why is it a permutation problem? Plus why do we use 5! × 3! in a permutation problem, isn't it used in combinations? as 5!X 3! shows the nos of ways it can be arranged 'in any order' So how can we use that in a permu where order DOES matter!

The reason I think its a permutation problem is because of http://answers.yahoo.com/question/index?qid=20110530100750AAXrtRp
other than that I don't relly know why it is a permutation problem even.

 

[Dick]

Draw a picture. How many ways can you position a block of 5 boys in a line of 8 people with 3 girls? Then given that number of patterns how many ways can you shuffle the boys and girls around in their positions?

 

[mutineer123]

Draw a picture. How many ways can you position a block of 5 boys in a line of 8 people with 3 girls? Then given that number of patterns how many ways can you shuffle the boys and girls around in their positions?

I don't think you got what I was trying to say. Yes i know 'how' it is 5! X 3! ways, but i don't know 'why' it is a permutation, because they are clearly being selected randomly...

 

[Dick]

I don't think you got what I was trying to say. Yes i know 'how' it is 5! X 3! ways, but i don't know 'why' it is a permutation, because they are clearly being selected randomly...

Ok, so you know the solution. If it's any help, I would not have written 4C1 instead of 4. The answer is just 4*5!*3!. Why you would write 4C1 instead of 4 escapes me. Is that what you are asking?

 

[mutineer123]

Ok, so you know the solution. If it's any help, I would not have written 4C1 instead of 4. The answer is just 4*5!*3!. Why you would write 4C1 instead of 4 escapes me. Is that what you are asking?

No, ok firstly, do you agree its a permutation sum and NOT a combination?

 

[Dick]

No, ok firstly, do you agree its a permutation sum and NOT a combination?

I agree that it has somewhat more to do with permutations, but I also don't think there is an absolutely clear separation between permutation problems and combination problems. Sometimes you have to use a little of both.

 

[zooxanthellae]

I hesitate to do this since I'm not well-versed in combinations/permutations, but don't we have more than 4(5!)(3!)? The 5! comes from the block of boys, and the 3! from a block of girls, but aren't there more different combinations of girls around the boys? 1 girl in front, 2 in back, 2 in front 1 in back, 3 in front 0 in back. I'm wondering myself how those are accounted for (not to hijack the thread...).

 

[mutineer123]

okay, il use that logic in the next problems i solve. Thanks Dick, and btw if you have time could you look at another problem that I am having trouble with? https://www.physicsforums.com/showthread.php?p=3831083&posted=1#post3831083

 

[azizlwl]

I hesitate to do this since I'm not well-versed in combinations/permutations, but don't we have more than 4(5!)(3!)? The 5! comes from the block of boys, and the 3! from a block of girls, but aren't there more different combinations of girls around the boys? 1 girl in front, 2 in back, 2 in front 1 in back, 3 in front 0 in back. I'm wondering myself how those are accounted for (not to hijack the thread...).

You forget another one, 3 in the back 0 in front.
<--
GBGG or 3!5!
GGBG or 3!5!
GGGB or 3!5!
BGGG or 3!5!
=4.3!5!

Anymore combination you can add?

 

[zooxanthellae]

You forget another one, 3 in the back 0 in front.
<--
GBGG or 3!5!
GGBG or 3!5!
GGGB or 3!5!
BGGG or 3!5!
=4.3!5!

Anymore combination you can add?

Ah, got it. Thanks.