Seven children — A, B, C, D, E, F, and G — are going to sit in seven c

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Club C and F as one, leaving us with six children. Total arrangements = 6! and since CF and also sit as FC, multiply 6! by 2 = 1440. Answer C.
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Hi All,

If you have trouble "seeing" how all of the possible options occur, it sometimes helps to do a small "sample" so that you can spot the overall pattern.

We're told that there are 7 children sitting in a row. We're also told that children "C" and "F" must sit next to one another. We're asked for the total possible number of arrangements.

Let's start by putting "C" and "F" right at the beginning...

C F _ _ _ _ _

The remaining 5 children could be arranged in any possible order...

C F 5 4 3 2 1

So there are 5! = 120 options with C "first" and F "second"

F COULD be the first one though.....

F C 5 4 3 2 1

So there are ANOTHER 120 options in this set-up.

Nothing states that C and F have to be in the first 2 chairs though....We could have....

_ C F _ _ _ _

Here, we still end up with 5! though....

5 C F 4 3 2 1

So that's another 120 options and we get another 120 if we have...

5 F C 4 3 2 1

This pattern ultimately proves that there are 240 options for every set of spots where C and F could sit. Since there are 6 of those....

* * _ _ _ _ _
_ * * _ _ _ _
_ _ * * _ _ _
_ _ _ * * _ _
_ _ _ _ * * _
_ _ _ _ _ * *

6(240) = 1440

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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I also used the "glue" method I learned in the Manhattan guides.

This means that we consider the constraint (in our case the 2 children that should sit next to each other, as one unit).
So, instead of 7! we have 6! = 720.

Since there are 2 children, one of them can sit first and the other second, or the oter way around. For this reason, we multiply 720*2 = 1400, which gives us ANS C.

A "twist" in this problem could be that the 2 children should not sit next to each other. I believe that in this case we would calculate 7! = 5040 (the ways all 7 can be arranged). Then we would find 6! = 720 (and multiply by 2 for the same reason as above). However, in the end we would subtract the 2, leaving us with 5040 - 1440 = 3600.

Is this right or wrong...?

for such questions , we can treat both as one then total person=6..
these 6 can be arranged in 6! ways.. but within these 6 , one consists of two people who can be arranged in 2 ways CF or FC.. so ans =6!*2=1440
ans C
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+1 for C. 6!*2!=1440

MAGOOSH OFFICIAL SOLUTION:

First, we will consider the restricted elements — C & F. How many ways can they sit next to each other in a row of seven chairs? Well, first of all, how many “next to each other” pairs of chairs are there?
X X _ _ _ _ _
_ X X _ _ _ _
_ _ X X _ _ _
_ _ _ X X _ _
_ _ _ _ X X _
_ _ _ _ _ X X

There are six different pairs of “next to each other” chairs. For each pair, children C & F could be in either order, so that’s 6*2 = 12 possibilities for these two.

Now, consider the other five children. For any configuration of C & F, the remaining five children could be seated in any order among the five remaining seats. Five items in any order — that’s a permutation of the 5 items —- 5P5 = 5! = 120. For any single configuration of C & F, there are 120 ways that the other children could be seated in the remaining seats.

Finally, we’ll combine with the Fundamental Counting Principle. We have 12 ways for the first two, and 120 ways for the remaining five. That’s a total number of configurations of 12*120 = 1440.

Answer = C
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treat C and F as one entity . So in total we have 6 people to be seated
6 people can be seated in 6 ! ways = 720 ways
we can see that C and F can arrange among themselves in 2 ! ways = 2 ways
Total ways = 720 * 2 = 1440 ways
correct answer - C

We have total 7 seats to accommodate 7 children.C & F are supposed to sit together.Let's consider C&F a single entity..So we need to find out 6 places for 6 children(6!).Also C & F can be sit in 2! ways..No of configurations will be 6!*2!=1440