Permutation problem

[ceeb]

I am trying to find the number of ways to arrange a classroom in 'boy, girl' order, there are 6 boys and 6 girls, I know I need to find [MATH]6![/MATH] to find how many ways the girls can be arranged and do that again for the boys, so now I have 720 ways the girls can be arranged and 720 ways the boys can be arranged.

But I read on a website that these two values must be multiplied. I have no idea why.

 

[Dr.Peterson]

Please state the exact wording of the problem as given to you. These problems are very sensitive to details.

But if the problem is that there are 6 seats for the girls and 6 for the boys, then for each of 6! ways you place the girls, there will be 6! ways to place the boys. Therefore you multiply.

 

[pka]

I am trying to find the number of ways to arrange a classroom in 'boy, girl' order, there are 6 boys and 6 girls, I know I need to find [MATH]6![/MATH] to find how many ways the girls can be arranged and do that again for the boys, so now I have 720 ways the girls can be arranged and 720 ways the boys can be arranged. But I read on a website that these two values must be multiplied. I have no idea why.

Please carefully read Prof Peterson question & please do answer. I will try to address that.
If we have twelve chairs in a line and it does say boy-girl, we start with a boy and place the boys in every other chair That can be done in \(\displaystyle 6!\) ways. Now we place a girl into each empty chair. That can also be done in \(\displaystyle 6!\) ways. Your question is about why multiply those two.

Well every time we place the boys, then there are \(\displaystyle 6!=720\) ways to place the girls. For every one arrangement of boys there are \(\displaystyle 720\) arrangements of girls. The gives you \(\displaystyle 6!\cdot 6!\) total ways.

 

[ceeb]

Please state the exact wording of the problem as given to you. These problems are very sensitive to details.

But if the problem is that there are 6 seats for the girls and 6 for the boys, then for each of 6! ways you place the girls, there will be 6! ways to place the boys. Therefore you multiply.

Thank you Dr.Peterson! You were correct, the problem is as you described, the classroom layout is as follows.
"Each of the seats alternates between boy and girl. Find how many ways this can be done"

So I believe that makes you right in your justification
Thanks again for you help!

 

[Dr.Peterson]

No, I think this changes the problem considerably, because we no longer have a single line that alternates: BGBGBGBGBGBG. Instead, we have three sets of four, each of which can separately alternate, but with no necessary relation. So it could be BGBG BGBG BGBG, or BGBG GBGB BGBG, and so on.

What extra multiplications will you need in order to account for this?

Actually, now that I think about it, the problem as you wrote it didn't eliminate GBGBGBGBGBGB, did it? So we would have needed to double the answer.

Important lesson: Follow the rules we've given, and show the entire problem exactly as given, from the start, to avoid wasting our time and yours!

 

[pka]

As you can see I modified the diagram so the blue places are for the boys & the red are for the girls.
Now I still maintain that there are \(\displaystyle 6!\cdot 6!\) to arrange this group if boy-girl means on each side of a student-desk there is a boy to the left of a girl.
But if it means only that there is a boy & a girl on each of the six sides, order does not matter then the number of ways is \(\displaystyle 2^6\cdot 6!\cdot 6!\)

 

[Dr.Peterson]

I presume we still haven't seen the actual wording of the problem. Clearly, it is essential to know what is meant by "boy, girl order", which is very unclear, and hopefully is not all the problem said. I can think of several additional interpretations.