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Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to Child A on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children? (A) 240 (B) 480 (C) 720 (D) 1440 (E) 3600

Re: Seven children are going to sit in seven chairs in a row [#permalink]

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29 Jan 2013, 11:28

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This post received KUDOS

+1 A

Here, we have to use the glue-method: Suposse that BAG or GAB are just one child. Let's call it X. So, we have: X, C, D, E, F: Just 5 kids. We arrange them: 5! =120

Then, we have combinations like this: X - C -D - E -F C - X - D - E - F etc...

Re: Seven children are going to sit in seven chairs in a row [#permalink]

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29 Jan 2013, 07:14

Arrange B A and G in the following way --> BAG---- (4 open slots available after BAG). The total number of ways of arranging while fixing BAG in their spots is 4!. Since B and G can be interchanged we have 4!x2 number of ways having B A and G occupying the first 3 slots. Since there are 5 different ways to have A in the middle of B and G the total number of ways of arranging the children is (4!x2) x 5 = 240 (A)

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24 Jul 2013, 13:48

Question states as following, thus I am not able to understand why others are considering GAB as a possible arrangement. "Child A has to sit next to both B & G, with these two children immediately adjacent to Child A on either side."

As per my understanding I am able to figure out following possible arrangements.

[BGA] C D E F = 5! permutations. [ABG] C D E F = 5! [GBA] C D E F = 5! [AGB] C D E F = 5!

120 x 4 = 480 arrangements, I got this, is it right ?
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Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Well, when I post a question, the GC system makes me guess on the difficulty ---- take that as a guess, no more. It's true, if you are fluent combinatorics and counting problems, this problem is a breeze, but folks who are not particular adroit at counting might start, for example, listing all possibilities. All math is impossibly difficult when you don't see how to do it, and trivially easy when you do, and that's doubly true for counting problems. I was basing my guess of the difficulty on the frequency of questions I see about such topics. If this question was easy for you, nave81, that may reflect more on your talent than on the nature of the question.

PiyushK wrote:

Question states as following, thus I am not able to understand why others are considering GAB as a possible arrangement. "Child A has to sit next to both B & G, with these two children immediately adjacent to Child A on either side." As per my understanding I am able to figure out following possible arrangements. [BGA] C D E F = 5! permutations. [ABG] C D E F = 5! [GBA] C D E F = 5! [AGB] C D E F = 5! 120 x 4 = 480 arrangements, I got this, is it right ?

Dear PiyushK, I regret to tell you, sir, that you are misreading the question. The wording here is tricky. Child A has to sit next to both B & G, with these two children immediately adjacent to Child A on either side. These "two children" are B & G, and each one of them is adjacent to A ---- thus BGA & AGB are not a possibilities because B is not adjacent to A, and ABG & GBA are not possibilities because G is not adjacent to A. The wording implies that B must be adjacent to A and that G must be adjacent to A, and that these two are on "either side" of A, that is on opposite sides of A. All four of your cases had B adjacent to G, not a requirement, and had them both on the same side of A, which contradicts the condition given in the text. That's an example of a math idiom --- "on either side of A" --- this means that the two items in question, B & G, cannot possibly be on the same side of A; they must be on opposites sides of A. Does all this make sense? Mike
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23 Aug 2014, 06:20

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30 Jun 2016, 01:39

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