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Seven children are going to sit in seven chairs in a row

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Seven children are going to sit in seven chairs in a row [#permalink] New post 28 Jan 2013, 14:14
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76% (02:07) correct 24% (01:20) wrong based on 108 sessions
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


For an in-depth discussion of counting questions such as this one, as well as the complete solution to this question, see:
http://magoosh.com/gmat/2013/gmat-count ... trictions/
[Reveal] Spoiler: OA

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Re: Seven children are going to sit in seven chairs in a row [#permalink] New post 28 Jan 2013, 14:37
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First we can imagine the trio ABG as one item to arrange along with four other items: C , D, E, F.

This can be done in 5! = 120 ways

However the trio can be arranged in 2 ways: BAG or GAB.

So, total number of ways to sit the seven children is equal to 2*120 = 240 (option A)

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Re: Seven children are going to sit in seven chairs in a row [#permalink] New post 29 Jan 2013, 06:14
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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Re: Seven children are going to sit in seven chairs in a row [#permalink] New post 29 Jan 2013, 10:28
+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...

But X can be B-A-G or G-A-B
Then: 120 * 2 = 240

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Re: Seven children are going to sit in seven chairs in a row [#permalink] New post 02 Feb 2013, 04:00
Is this really a 700 question? seemed too easy :?
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Re: Seven children are going to sit in seven chairs in a row [#permalink] New post 24 Jul 2013, 12:48
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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Re: Seven children are going to sit in seven chairs in a row [#permalink] New post 24 Jul 2013, 13:05
Expert's post
nave81 wrote:
Is this really a 700 question? seemed too easy

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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Re: Seven children are going to sit in seven chairs in a row [#permalink] New post 23 Aug 2014, 05:20
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Re: Seven children are going to sit in seven chairs in a row   [#permalink] 23 Aug 2014, 05:20
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