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There are 3 couples sitting in a row (1 couple consists of 1 boy and 1 girl).
a. How many different sitting arrangements are possible?
b. What if one boy and one girl of a couple don't want to sit next to each other?

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Schoolboy wrote:
There are 3 couples sitting in a row (1 couple consists of 1 boy and 1 girl).
a. How many different sitting arrangements are possible?
b. What if one boy and one girl of a couple don't want to sit next to each other?
a.

I would use the slot method: ___ ___ ___ ___ ___ ___ (our 6 slots)
Slot 1: 6 potential people can go in this slot
Slot 2: Only 1 person can go in this slot (the partner to the person who goes in slot 1)
Slot 3: 4 potential people can go in this slot
Slot 4: Only 1 person can go in this slot (the partner to the person who goes in slot 3)
Slot 5: 2 potential people can go in this slot
Slot 1: Only 1 person can go in this slot (the partner to the person who goes in slot 5)

Arrangements = 6*1*4*1*2*1 = 48

b.

Method --> do part a. and then subtract the arrangements where the boy and girl who do not want to sit together, actually sit together.

Ways this can happen:

(a) XBGXXX
(b) XGBXXX
(c) XXXBGX
(d) XXXGBX

For each of the above scenarios there are 2 ways you can arrange the seating plan. Let's use the slot method to examine (a):

Slot 1: 1 person can go in this slot (the partner of B)
Slot 2: 1 person (aka B)
Slot 3: 1 person (aka G)
Slot 4: 1 person can go in this slot (the partner of G)
Slot 5: 2 potential people can go in this slot (either the boy or the girl of the last couple)
Slot 1: Only 1 person can go in this slot (the partner to the person who goes in slot 5)

Arrangements = 1*1*1*1*2*1 = 2

If you do this for (a), (b), (c) and (d) you'll see that there are two arrangements for each scenario. Thus the number of arrangements to subtract from 48 is 8.

The answer is 48 - 8 = 40.

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Well, the answer to your first question is fairly easy. We have three couples that can go in three locations:

C1, C2, C3

How many ways can they be organized? Well, initially we would need to select one couple to go far left. We have three choices. Then we would select the couple to go in the middle. We have two choices for that. Then the last couple would go in the far right location. So we would have 3x2x1 or 3! possibilities.

But that is not all. We must consider how the couples are arranged. We might have the man on the right and the woman on the left, or vice versa. So the number of possibilities are double.

From that number, we would need to calculate the number of ways that one boy and one girl might sit together and subtract those possibilities. They are:

xbgxxx
xgbxxx
xxxbgx
xxxgbx

So four.

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Elias Latour
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+1 (646) 736-7622

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Schoolboy wrote:
There are 3 couples sitting in a row (1 couple consists of 1 boy and 1 girl).
a. How many different sitting arrangements are possible?
b. What if one boy and one girl of a couple don't want to sit next to each other?
a. consider each couple as a unit, So there are 3! ways to seat the 3 couples.
Within each unit there 2 ways to sit the boy and girl, so need to multiply 3! by 2^3 equaling 48 ways to sit the 3 couples

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Here's another solution that some may find easier.

We have 6 empty seats and three couples. How many people can sit in the first seat? 6 people -- any of the people waiting to be seated. However, once that happens, the person sitting next to him or her must be the other half of the couple, so we have only one choice. I will simulate this with the following notation:

(6) (1)

Once those two are seated, how many people can be seated in the third seat? 4. But once that person is seated, again, only one person can sit next to him or her. So we can notate that as:

(6) (1) (4) (1)

Finally, in the fifth seat, 2 people can be seated. So we have:

(6) (1) (4) (1) (2) (1)

And all we have to do is multiply the numbers together.

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Elias Latour
Verbal Specialist @ ApexGMAT
blog.apexgmat.com
+1 (646) 736-7622

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