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

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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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)

+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

Kudos, if you like it

Is this really a 700 question? seemed too easy

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 ?

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.


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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