Arranging friends, permutations, have the answer, not sure on some parts

[mr_coffee]

Hello everyone.
The book has this problem:

(h) You are arranging six of your friends Alice, Bob, Charles, Diana, Francine, and George, in a row so that you can take their picture.

(i). Alice and Bob have had a fight and refuse to stand next to each other. How many ways are there to arrange your six friends?

Well I took the compliment of this problem I think its called and said, how many ways are there to make sure alice and bob stand next to each other so you have:

[A,B],C,D,F,G = 5!
You can arrange this in 5! + 5! ways, because, the first arrangement is shown, and the next arrangment would be:
[B,A],C,D,F,G = 5!

This is 120 + 120 = 240 ways to keep them together
They can fail to stand next to each other in 6! ways, so
6! - 240 = 480 which is what the book had.

Now the 2nd part:
(ii) A 7th friend Elivs, arrives, and Alice and Bob both insist on standing next to Elivs. How many ways are there to arrrange your seven friends?
Well this one I thought I could appy the same way as the first part i wrote:

[A,B,E],C,D,F,G
Okay now you still have 5! ways to arrange them in that order
and you can arrange [A,B,E] 3!
So i thought it would be 3! + 5! + 5! + 5! = 366, but the answer is 240.


(iii)
Alice and Bob have made up and insist on standing next to each other. ANd you have just realized that you can't place Elvis at either end of the row or he will fall over becuase he's dead. Now how many ways are ther to arrange your seven friends?

Welll if A and B stand together and E can't be at either end I thought maybe this would work:
[A,B],C,D,F,G

now E can't go inbetween A and B because they want to stand near each other, and it can't be after G
so it has to be in Either:
[E,C,D,F] and there are 4! to arrange them
and:
[A,B],[E,C,D,F,]G
But now this is getting messy... would it be like 3! + 4! + 5! which is wrong the book got 960
Any ideas how they got 240? or 960 and did I find part (i) correctly or just lucky?



THanks

 

[drpizza]

1st part seems good.

For the second part, if it's ABE, is A next to E?
You have the right idea, but E would need to be between A and B for them to both be next to E. Since it's AEB or BEA, you've already done the work for this in the first part. (Treat AEB the same way you treated AB)

Also, remember where the factorial comes from: The counting principle.
Treat [AB] as one entity (then multiply by 2 at the end to include [BA]
You've got 6 things to arrange, but 1 of those things can't be at either end. How many choices do you have for the 1st position? How many choices remaining for the 2nd position? How many choices for the 3rd position...

 

[calcnd]

I have no idea how to do permutations properly, so my answer is obviously incorrect.

I go a simplified expression of

$$16[2!(4!)] = 768$$

I'd be interested to learn how to do this properly.

 

[mr_coffee]

Thanks guys i got the answers now:

For part (ii) the answer is 2 x 5! = 240.

For part (iii) the answer is 2 x 6! - 2x2x5! = 960

i'm still alittle iffy on (iii) but it seems like that works. If anyone wants to add anything for the last one they are welcome!
thanks!

 

[JasonRox]

Thanks guys i got the answers now:

For part (ii) the answer is 2 x 5! = 240.

For part (iii) the answer is 2 x 6! - 2x2x5! = 960

i'm still alittle iffy on (iii) but it seems like that works. If anyone wants to add anything for the last one they are welcome!
thanks!

Number 3 works like this...

This is how I do it, and I'm sure it will help you.

To get the first part, assume the two sit together. Forget about the dead guy. How many ways can you arrange them?

Well, let the couple be one object (since they have to sit together), but now we have 6 objects left, which has 6! ways of arranging 6 objects. Now, one object consists of two objects and that has 2! ways of arranging them. So, now we have 2!x6!=2x6!.

Now, let's look at the second part. We have to do this because we must subtract all those where the dead guy sits the ends. We didn't include this in the first part.

So, count how many way the dead guy can sit at one end of the table and the couple as an object, like above. So, let's count with the dead guy at the front. Like D represent the dead guy, and AB the couple, and O the other people.

So, we are arranging this...

D(AB)OOOO

So, we have to let D sit in the first slot because we want to count all the permutations of D being at the front. How many is there? Well, there are 6 objects left. Now, again put the couple as an object, so that's 5 objects. So, 5! for them 5. Now, the couple has 2! ways of being arranged (like above), so we now have 2x5!.

Now, we did all this with D at the front, but now let's do it with D at the back, which is the same, and so we get 2x2x5!

We subtract this number from 2x6! because this number includes all possible arrangements, which has D at the front and back, but we have to subtract those out because that's the question.

So, we have 2x6!-2x2x5!.

I hope that helped.

 

[mr_coffee]

Thanks janson! I got lost on one part,

When you said:

Now, we did all this with D at the front, but now let's do it with D at the back, which is the same, and so we get 2x2x5!

I understand how you goet 2x5! once you let the dead guy be fixed in the front, and you treat AB as 1 object, thus 2x5! was to arrange them.

But if you stuck the dead guy in the back you would get somthing like this:
(AB)OOOOD, same thing, 2x5! so, if you have in the first case 2x5! and now in the second case 2x5! wouldn't u get (2x5!)(2x5!) or 2x2x5!5!
? I know my answer is wrong and yours is right but if you have 2x5! ways to arrange them in both cases, why do you only carry over the other 2, getting 2x2x5! instead of the 2x5! ?

thanks :)

 

[JasonRox]

Thanks janson! I got lost on one part,

When you said:


I understand how you goet 2x5! once you let the dead guy be fixed in the front, and you treat AB as 1 object, thus 2x5! was to arrange them.

But if you stuck the dead guy in the back you would get somthing like this:
(AB)OOOOD, same thing, 2x5! so, if you have in the first case 2x5! and now in the second case 2x5! wouldn't u get (2x5!)(2x5!) or 2x2x5!5!
? I know my answer is wrong and yours is right but if you have 2x5! ways to arrange them in both cases, why do you only carry over the other 2, getting 2x2x5! instead of the 2x5! ?

thanks :)

No, we don't multiply.

We have the "or" rule here. If the D is in the front OR back, so we have...

2x5! + 2x5! = 2x2x5!

 

[mr_coffee]

Oooo my bad, thanks again!