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Re: Permutation Combination bookshelf [#permalink]
27 Nov 2009, 15:01

Expert's post

1) each French book has to be separated at least by one English book, so we have following order: *FE*FE*FE*FE*FE*FE*FE*FE*F* where * denotes 10 possible places for remained 11-8=3 English books

2) Now, let's count in how many ways we can place remained English books:

C^{10}_3 - all 3 books at distinct places. P^{10}_2 - 2 books together and remained book at distinct place (order is important). C^{10}_1 - all 3 books together.

maybe there is a better solution
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Re: Permutation Combination bookshelf [#permalink]
27 Nov 2009, 16:45

Bunuel, I wish I could adopt your way to look at the problem. Sometimes, it's an easy one lost in translation. Walker, though your approach is different, but it's pretty interesting. Thanks again guys, appreciate your help here.

Re: Permutation Combination bookshelf [#permalink]
28 Nov 2009, 21:23

I am not very clear guys. Buenel, in your solution you have only considered the different ways that the 9 french books can be placed in 12 position. However, even the English books can be placed in their respective slots in more than one way. i.e. let the English books be 1e, 2e, 3e etc. Now different placement of these english books can yield more possible combinations.

F 1e F 2e F 3e or F 3e F 1e F 2e or F 2e F 1e F 3e

Re: Permutation Combination bookshelf [#permalink]
29 Nov 2009, 13:23

It is true that English books may have different positions but that's not what the question is asking. It's only a matter of placing 9 books in 12 slots.

Re: Permutation Combination bookshelf [#permalink]
29 Nov 2009, 13:37

4

This post received KUDOS

Expert's post

oracle wrote:

I am not very clear guys. Buenel, in your solution you have only considered the different ways that the 9 french books can be placed in 12 position. However, even the English books can be placed in their respective slots in more than one way. i.e. let the English books be 1e, 2e, 3e etc. Now different placement of these english books can yield more possible combinations.

F 1e F 2e F 3e or F 3e F 1e F 2e or F 2e F 1e F 3e

Hope I am clear in expressing my doubt

Think I understand your point. You are saying that along with arrangements with no French books being adjacent, English and French books themselves could be arranged in different ways. But I don't think that this is the case. Though it's quite ambiguous question in a sense. Basically when GMAT wants us to consider some items as distinct it specifies this OR it's quite obvious. In original question we don't know whether these books are distinct or not: maybe all French and English books are the same, maybe not, we don't know that.

If the question were: how can we arrange 11 boys and 9 girls so that no girls are together, then the answer would be 12C9*11!*9!. As it's obvious that they are all different.

If the question were: how can we arrange 11 A-s and 9 B-s so that no B-s are together, then the answer would be 12C9. As it's obvious that A-s and B-s are the same.

Re: Permutation Combination bookshelf [#permalink]
29 Nov 2009, 14:11

11 english books can be arranged in 11 slots (say) in 11! ways. There will be 10 empty spaces between adjacent english books and another 2 empty spaces at either ends into which the french books can be placed such that no two french books are adjacent to one another. This is done in 12C9 ways.

Total number of ways should therefore be 12C9 * 11!.
_________________

Re: Permutation Combination bookshelf [#permalink]
29 Nov 2009, 14:56

6

This post received KUDOS

Expert's post

BarneyStinson wrote:

11 english books can be arranged in 11 slots (say) in 11! ways. There will be 10 empty spaces between adjacent english books and another 2 empty spaces at either ends into which the french books can be placed such that no two french books are adjacent to one another. This is done in 12C9 ways.

Total number of ways should therefore be 12C9 * 11!.

Not so.

Let's say we have A1, A2, B1, B2 (meaning that A-s and B-s are distinct). We want to arrange them so that no B-s are adjacent:

*A1*A2* and we can place B1 and B2 in 3 empty slots. It can be done in 3C2 # of ways. BUT A1 and A2 can be arranged like *A1*A2* OR *A2*A1*, plus B1 and B2 can be arranged as B1B2 or B2B1.

Total # of ways 3C2*2!*2!=12.

Still if not convinced: B1,A1,B2,A2 B1,A2,B2,A1 B2,A1,B1,A2 B2,A2,B1,A1

A1,B1,A2,B2 A2,B1,A1,B2 A1,B2,A2,B1 A2,B2,A1,B1,

B1,A1,A2,B2 B1,A2,A1,B2 B2,A1,A2,B1 B2,A2,A1,B1

BUT again this is the case when we have DISTINCT items. So, if we were told that all French book are different and all English books are different, then the answer would be 12C9*11!*9!.

In our original question we are not told that French books are different and are not told that English books are different. So # of ways would be 12C9.

Let's consider the easier example: # of ways to arrange two A-s and 2 B-s so that no B-s are adjacent: 3C2=3.

Re: Permutation Combination bookshelf [#permalink]
06 Dec 2009, 02:07

Thanks Buenel. Ya, It helps, here I was confused because it does not mention whether we should consider the English Books and the French books to be distinct, or as you said, like any A's and B's, which reduces the total number of possible arrangements.