Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be

Consider B and E as one entity. The arrangement of these two is possible in '2' ways

Now B&E as one entity and the other 4 children = total of 5 entities that need to be arranged = 5! ways

So, number of ways of arranging in which B and E sit together = 5! * 2

Total number of ways of arranging the 6 students = 6!

Total number of ways of arranging 6 students together - Number of ways in which B and E sit together = Total number of ways in which B and E don't sit together.

6! - ( 5! * 2)
= (6 * 5!) - (2 * 5!)
= 5! (6 - 2)
= 5! * 4
= 120 * 4
= 480 ways

Hence, B

Hi All,

If you find the explanations to certain prompts to be a bit too complex, then you might be able to break a prompt down into smaller 'pieces' (and look for patterns among those pieces) and still get to the correct answer.

Here, we're told that 6 children, (who I'll call A,B,C,D,E and F) are to be seated in a single row of six chairs. We're told that B cannot sit next to E. We're asked how many different arrangements of the six children are possible.

IF there were no 'restrictions', then the number of possible arrangements would be 6! = (6)(5)(4)(3)(2)(1) = 720. However, there IS a restriction (B cannot sit next to E), so we have to eliminate all of the possible arrangements that include B sitting next to E....

_ _ _ _ _ _

IF the first two spots were B and E, there would be....

B E 4 3 2 1 = 24 possible arrangements that start with BE.

In that same way, if the first two spots were E and B, there would be...

E B 4 3 2 1 = 24 possible arrangements that start with EB.

Notice how it's 24 and 24? What happens when B and E are in the second and third spots.....?

4 B E 3 2 1 = 24 possible arrangements. Using the pattern we've proven (above), there would be ANOTHER 24 arrangements when we switched the B and E...

4 E B 3 2 1 = 24 possible arrangements. This is clearly a pattern now, so we just have to count up all of the '24s'

1st/2nd spots = 24+24
2nd/3rd spots = 24+24
3rd/4th spots = 24+24
4th/5th spots = 24+24
5th/6th spots = 24+24

Total = 24(10) = 240 arrangements that would have B and E next to one another. We have to REMOVE those from the total...

720 - 240 = 480 possible arrangements that fit what we're looking for.

Final Answer:
GMAT assassins aren't born, they're made,
Rich

We are given that Arya, Betsy, Chen, Daniel, Emily, and Franco are to be seated in a single row of six chairs, and that Betsy cannot sit next to Emily. We need to determine how many different arrangements of the six children are possible.

We can use the following equation:

[Total number of arrangements] = [number of arrangements with Betsy next to Emily] + [number of arrangements with Betsy not next to Emily].

Let’s rearrange the above equation as:

[Number of arrangements with Betsy not next to Emily] = [total number of arrangements] - [number of arrangements with Betsy next to Emily].

Since we are arranging 6 children, they can be arranged in 6! = 720 ways. Now we can determine how many ways to arrange them when Betsy is next to Emily. We can represent this arrangement as:

[A]-[B-E]-[C]-[D]-[F]

Notice that since Betsy is next to Emily, we consider them as 1 unit. Since there are 5 units to arrange, those units can be arranged in 5! = 120 ways. We also must account for the fact that when sitting together, Betsy and Emily can be arranged in 2! ways ([B-E] or [E-B]), so the total number of ways to arrange the children when Emily is next to Betsy is 2 x 120 = 240 ways.

Thus, the number of ways to arrange the children when Betsy is not next to Emily is 720 - 240 = 480 ways.

Answer: B

Hi Brent,

I used another approach you teach. Seems easier.

Started with the most restrict: (1* x 4 x 4 x 3 x 2 x 1) = 96 possibilities. *means Betsy or Emily. Then, multiplied by 5 different chairs they could be seated 96 x 5 = 480.

Make sense? Did I miss something?

Thanks!

That seems right. However, to be sure, can you tell me what each value represents?

Cheers,
Brent

That seems right. However, to be sure, can you tell me what each value represents?

Cheers,
Brent[/quote]

Let's start with Betsy. This is: 1
Then we could have any of the other: 4 (not Emily)
Next: Any other 3 + Emily: 4
Next: Any other 2 + Emily: 3
Next: The other + Emily: 3
Next: Emily or the one not chosen before: 1

Let's start with Betsy. This is: 1
Then we could have any of the other: 4 (not Emily)
Next: Any other 3 + Emily: 4
Next: Any other 2 + Emily: 3
Next: The other + Emily: 3
Next: Emily or the one not chosen before: 1[/quote]

Is your first step seating Betsy in one of the 5 seats, or is it selecting someone for seat #1?

Cheers,
Brent

Is your first step seating Betsy in one of the 5 seats, or is it selecting someone for seat #1?

Cheers,
Brent[/quote]

It would be seating Betsy. That way I really know the next couldn't be Emily. When I multiply by 5 at the end it is counting all the possibilities that I could start.

Tks!!

I think I am finally getting it

So we have two people who cannot sit together:

There are 10 ways in which they can sit 2 people together so we multiple the probability of picking one after another by 10 -> (2/6 * 1/5) which gives us 20/30 and then we multiple this by 6! to get 480.

All the other ways listed above make no sense to me.

­Note: forgot the 2! ways they can sit withing themselves.