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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 eachother. 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 eachother 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 eachother 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 eachother. 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 eachother, 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

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

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