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Intern  Joined: 24 Sep 2009
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Two couples and one single person are seated at random in a row of fiv  [#permalink]

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7 00:00

Difficulty:   95% (hard)

Question Stats: 41% (02:42) correct 59% (02:28) wrong based on 627 sessions

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Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2

Originally posted by benjiboo on 25 Sep 2009, 13:07.
Last edited by Bunuel on 18 Jun 2020, 08:24, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Two couples and one single person are seated at random in a row of fiv  [#permalink]

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1
38
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2

Let's find the opposite probability and subtract it from 1.

Opposite event that neither of the couples sits together is event that at leas one couple sits together. # of arrangements when at leas one couple sits together is sum of arrangements when EXACTLY 2 couples sit together and EXACTLY 1 couples sit together.

Couple A: A1, A2
Couple B: B1, B2
Single person: S

EXACTLY 2 couples sit together:
Consider each couple as one unit: {A1A2}{B1B2}{S}, # of arrangement would be: $$3!*2!*2!=24$$. 3! # of different arrangement of these 3 units, 2! arrangement of couple A (A1A2 or A2A1), 2! arrangement of couple B (B1B2 or B2B1).

EXACTLY 1 couple sits together:
Couple A sits together: {A1A2}{B1}{B2}{S}, # of arrangement would be: $$4!*2!=48$$. 4! # of different arrangement of these 4 units, 2! arrangement of couple A (A1A2 or A2A1). But these 48 arrangements will also include arrangements when 2 couples sit together, so total for couple A would be $$48-24=24$$;

The same for couple B: {B1B2}{A1}{A2}{S}, # of arrangement would be: $$4!*2!=48$$. Again these 48 arrangements will also include arrangements when 2 couples sit together, so total for couple B would be $$48-24=24$$;

$$24+24=48$$.

Finally we get the # of arrangements when at least one couple sits together is $$24+48=72$$.

Total # of arrangements of 5 people is $$5!=120$$, hence probability of an event that at leas one couple sits together would be $$\frac{72}{120}=\frac{3}{5}$$.

So probability of an event that neither of the couples sits together would be $$1-\frac{3}{5}=\frac{2}{5}$$

Hope it's clear.
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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7
4
I think I found a better method by using the "glue strategy" where you take 4 instead of 5 people indiating that 2 always sit next to each other. (In the end you have to multiply by two though)
Combinations for one couple sitting next to each other4!=24x2=48
Combinations for the other couple sitting next to each other 4!=24x2=48
overall 96

Watch out!
P(A or B) = P(A) + P(B) - P(A and B)
But now we P(A and B) to often and we have to delete it once

To calculate P(A and B) we glue both couples together and multiply in the end by 4
3!x4=24
96-24=72
1-(72/120)=2/5

What do you guys think?
##### General Discussion
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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4
1
total outcomes 5!
outcomes of couple A sit together: 2*4!, outcomes of couple B sit together: 2*4!, but we must deduct the possibility the 2 couples sit together: 3!*4
So outcomes of at least one couple sit together= 48+48-24=72
Outcomes of no couple sit together: 5!-72=48
possibility of no couple sit together: 48/5!=2/5
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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1
I used the glue method as referenced in Mgmat book

Total number of rearranging 5 ppl in 5 chairs = 5! = 120 (denominator)

Now lets resolve for the numerator:
A1 A2 C B1 B2
treat A1 and A2 as a single person and B1 and B2 as a single person...then rearranging 3 ppl in 3 positions is 3! = 6

we need to account for 8 variations in the seating arrangements:
A1 A2 C B1 B2
A1 A2 C B2 B1
A2 A1 C B1 B2
A2 A1 C B2 B1
B1 B2 C A1 A2
B1 B2 C A2 A1
B2 B1 C A1 A2
B2 B1 C A2 A1

Therefore the result is (8*3!)/5! = 2/5.
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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benjiboo wrote:

Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

1/5
1/4
3/8
2/5
1/2

I am giving my solution below which needs a little bit of thought but minimum case evaluations. The logic I use here is the one we use to solve SETS questions. Let me explain.

There are two couples. I don't want either couple to sit together. I would instead like to work with 'making them sit together'.

Would you agree that it is easy to find the number of arrangements in which both couples are sitting together? It is. We will work on it in a minute. Let me think ahead now.

How about 'finding the number of ways in which one couple sits together?' Sure we can find it but it will include those cases in which both couples are sitting together too. But we have already found the number of ways in which both couples sit together. We just subtract that number from this number and get the number of ways in which ONLY one couple sits together. Think of SETS here.

Let's do this now.

Number of arrangements in which both couples sit together:
Say the couples are C1 and C2. I try and arrange the loner. He can take positions 1, 3 and 5. He can sit at 3 places. For each one of these positions, the couples can switch their places, C1 can switch places within themselves and C2 can switch places within themselves. So number of arrangements such that both couples are together are 3*2*2*2 = 24

Number of arrangements such that C1 is together:
C1 acts as one group. Arrange 4 people/groups in 4! ways. C1 can switch places within themselves so number of arrangements = 4! * 2 = 48
But this 48 includes the number of arrangements in which both couples are sitting together.
So number of arrangements such that ONLY C1 sits together = 48 - 24 = 24
Similarly, number of arrangements such that ONLY C2 sits together = 24

Total number of arrangements = 5! = 120
At least one couple sits together in 24 + 24 + 24 = 72 arrangements
No couple sits together in 120 - 72 = 48 arrangements

Probability that no couple sits together = 2/5

(Ideally, you should see that 24 is 1/5th of 120 so you immediately arrive at 2/5)
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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3
2
Total ways to arrange = 5! = 120
Ways in which 1st couple can sit together = 4! X 2! = 48
Ways in which both couple sit together = 3! X 2! X 2! = 24
Ways in which ONLY 1st couple can sit together = 48-24 = 24 = Ways in which ONLY 2nd couple sit together
Ways in which neither couple sits together = Total ways – Ways in which both sit together –Ways in which only 1st couple sits together – Ways in which 2nd couple sits together
= 120-24 –24-24 = 120-72 = 48
Probability = 48/120 = 2/5
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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ok if we assume that each couples as a one person total 3 person in a row and the probablity of three person,which couples seat together, sit on five chairs is 3/5. But the question is asking neither of couples seat together so the opposite way 1-3/5=2/5

Keep It Stupid Simple
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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arnaudl wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2

I think the best combinatorial solution was given by heyholetsgo. Neat and short!

I didn't see a probabilistic approach, so I tried to work out one. Here it is:

There are 5 chairs, let's visualize them _ _ _ _ _
There is one single person, let's denote him/her by S.
S can choose from 5 chairs where to sit. There are some symmetrical placements for S:
1) S _ _ _ _ and _ _ _ _ S (0, 1, or 2 couples can be placed together)
2) _ S _ _ _ and _ _ _S_ (0 or 1 couple can be placed together)
And there is the third case:
3) _ _ S _ _ (0, 1, or 2 couples can be placed together).

So, let's compute for each case the probability that 0 couple sits together.

1) S _ _ _ _ or _ _ _ _ S:
$$\frac{2}{5}$$ probability that S sits on either chair 1 or chair 5.
Now start placing people on the remaining four chairs and express the corresponding probabilities:
$$\frac{4}{4}$$ for the first chair in the sequence of four remaining chairs
$$\frac{2}{3}$$ for the second chair (cannot be the first person's mate)
$$\frac{1}{2}$$ for the third chair (cannot be the mate of the second person)
$$\frac{1}{1}$$ for the last person to be placed
In conclusion, the probability of no pairs sittings together when S sits on either chair 1 or 5 is given by
$$\frac{2}{5}*\frac{4}{4}*\frac{2}{3}*\frac{1}{2}*\frac{1}{1}=\frac{2}{15}.$$

2) _ S _ _ _ or _ _ _ S _
Again, two possibilities for S, so $$\frac{2}{5}$$ probability for S to sit on either chair 2 or 4.
Now start placing people on the remaining four chairs and express the corresponding probabilities.
Treat only the _ S _ _ _ case, the other one has the same probabilities:
$$\frac{4}{4}$$ for the first chair from the left
$$\frac{2}{3}$$ for the second chair (cannot be the first person's mate because then the last two chairs will be occupied by the other couple)
$$\frac{1}{2}$$ for the third chair (cannot be the mate of the second person)
$$\frac{1}{1}$$ for the last person to be placed
In conclusion, the probability of no pairs sittings together when S sits on either chair 2 or 4 is given by
$$\frac{2}{5}*\frac{4}{4}*\frac{2}{3}*\frac{1}{2}*\frac{1}{1}=\frac{2}{15}.$$

3) _ _ S _ _
Here now we just have one possibility to place S, so $$\frac{1}{5}$$ probability for S to sit on chair 3.
Starting from the left, place people on chairs:
$$\frac{4}{4}$$ for the first chair from the left
$$\frac{2}{3}$$ for the second chair (cannot be the first person's mate)
$$\frac{2}{2}$$ for the third chair (we are already assured that we separated the two couples)
$$\frac{1}{1}$$ for the last person to be placed
In conclusion, the probability of no pairs sittings together when S sits on chair 3 is given by
$$\frac{1}{5}*\frac{4}{4}*\frac{2}{3}*\frac{2}{2}*\frac{1}{1}=\frac{2}{15}.$$

In conclusion, the requested probability is $$3*\frac{2}{15}=\frac{2}{5}.$$

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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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arnaudl wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2

Instead of doing so many calculations, why not straight up find the probability that no one sits together?

Sure, you can do the venn diagram approach and find individual circles and subtract the overlap, but this way was a bit easier for me to understand:

_ _ _ _ _ = 5 positions

The single person can sit in 3 spots (Later we will account for all the permutations by multiply by 3! = 6)
S _ _ _ _ (sitting at the end)
_ S _ _ _ (Sitting 1 spot from the end)
_ _ S _ _ (sitting in the middle)

Note the remaining options (_ _ _ S _ and _ _ _ _ S) are just inverses of the first two.

So let's look at the first one: S _ _ _ _

S A1 B1 A2 B2
S A1 B2 A2 B1

S B1...
S B2...

If we pick A1 for that 2nd spot, then the 3rd spot cannot be A2 since A's cannot since next to each other. That 3rd spot must be one of the 2 B's: B1 or B2. Then the 4th spot must be the remaining A2. Then the last spot will be the remaining B.

Note that A and B must alternate - NO MATTER where S is located.

This is true for the other 2 forms (_ S _ _ _) as well as (_ _ S _ _ ) - A's and B's must alternate.

So back to S _ _ _ _

That second spot S (_) _ _ _ has 4 options to choose from (A1, A2, B1, and B2). Whichever you choose for that second spot, the third spot you will have 2 options to choose from.

If you choose A1 for the 2nd spot, then the 3rd spot has 2 options (B1 or B2). The remaining 4th and 5th spots are self-chosen so there is no need to calculate those last spots in this case.

S A1 B1 A2 B2
S A1 B2 A2 B1

So here's the calculation:

(Out of 4 options A1, A2, B1, and B2- choose 1) * (Out of remaining 2 options, choose 1)

= (4C1) * (2C1) = 8

Multiply by 3! for each of the positions that "S" can occupy and all the permutations around it (including S _ _ _ _ and _ _ _ _ S, etc)

So we have 8 * 3! = 8 * 6 = 48 possibilities that no couples sit together

Since there are a total of 5 positions, the total possibilities is 5! = 120
You can also think of this as (5C1 * 4C1 * 3C1 * 2C1 * 1C1) = 5! = 120

So divide the 48 possibilities that no couples sit together by the total 120 possibilities and you get:

48/120 = 24/60 = 4 / 10 = 40%
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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3
1
Here the word to CATCH is NEITHER.

After the introduction of this word in question , the valid combination consist of all combination where there is NO couple are sitting together.

which means the required probability is -

1- ( Probability that BOTH couples are sitting together + Probability that ONE couple sits together)

Probability that BOTH couples sits together -
Consider couples as single entity , so we have three distinct entity which can be arranged in 3! ways and since couples can also be re-arranged in them selves so total valid arrangement is 3! * 2! *2! and total possible arrangement are 5!
Probability = 24/120 = 1/5

Probability that ANY ONE couple sits together -
now select one couple and arrange the effective 4 entity in 4!*2! = 48 , BUT this will contain the cases which have another couple together which we have already counted in case above so effective arrangements = 48-24 = 24
Similarly chose 2nd couple and effective arrangement will be 24
Probability = (24+24)/120 = 2/5

required probability = 1- (1/5+2/5)
= 2/5

Hope i was able to explain it!!!!

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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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ulas wrote:
ok if we assume that each couples as a one person total 3 person in a row and the probablity of three person,which couples seat together, sit on five chairs is 3/5. But the question is asking neither of couples seat together so the opposite way 1-3/5=2/5

Keep It Stupid Simple

Is this approach correct? It is similar to what I thought but didn't quite get it right

Cheers
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Posts: 65807
Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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jlgdr wrote:
ulas wrote:
ok if we assume that each couples as a one person total 3 person in a row and the probablity of three person,which couples seat together, sit on five chairs is 3/5. But the question is asking neither of couples seat together so the opposite way 1-3/5=2/5

Keep It Stupid Simple

Is this approach correct? It is similar to what I thought but didn't quite get it right

Cheers
J No, that's not correct. The probability that neither of the couples sits together in adjacent chairs does not equal to 1- {the probability that exactly 2 couples sit together}. It's 1 - {the probability that exactly 2 couples sit together} - {the probability that exactly 1 couple sits together}.

Also, the probability that EXACTLY 2 couples sit together is 24/5! not 3/5:
Consider each couple as one unit: {A1A2}{B1B2}{S}, # of arrangement would be: $$3!*2!*2!=24$$. 3! # of different arrangement of these 3 units, 2! arrangement of couple A (A1A2 or A2A1), 2! arrangement of couple B (B1B2 or B2B1).
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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mohan514 wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2

Responding to a pm:

I am guessing you are looking for a one line solution since the answer is a simple 2/5. I couldn't think of the logic that will help us arrive at the answer directly. I use sets to solve such questions. It's not very different from what Bunuel has used above.

Here is my solution:

This question isn't very different from our regular "5 people sit in a row such that A does not sit next to B. In how many ways is this possible?"
Here, there are 2 pairs of people who cannot sit next to each other. Hence, you need to take special care of the cases in which both sit with each other.

There are two couples. We don’t want either couple to sit together. Let’s go the reverse way – let’s make at least one of them sit together. We can then subtract this number from the total arrangements to get the number of arrangements in which neither couple sits together.

Would you agree that it is easy to find the number of arrangements in which both couples sit together? It is. We will work on it in a minute. Let’s think ahead for now.

How about ‘finding the number of ways in which one couple sits together?’ Sure we can easily find it but it will include those cases in which both couples are sitting together too. But we would have already found the number of ways in which both couples sit together. When we just subtract ‘both couples together’ number once from the total to avoid double counting, we will get the number of ways in which at least one couple sits together. Think of SETS here.

Let’s do this now.

Number of arrangements in which both couples sit together: Let’s say the two couples are {C1h, C1w} and {C2h, C2w} and the single person is S. There are three groups/individuals. They can be arranged in 3! ways. But in each couple, husband and wife can be arranged in 2 ways (husband and wife can switch places)

Hence, number of arrangements such that both couples are together = 3!*2*2 = 24

Number of arrangements such that C1h and C1w are together: C1 acts as one group. We can arrange 4 people/groups in 4! ways. C1h and C1w can be arranged in 2 ways (husband and wife can switch places).

Number of arrangements in which C1h and C1w are together = 4! * 2 = 48
But this 48 includes the number of arrangements in which C2h and C2w are also sitting together.
C2h and C2w also sit together in 48 ways (including the number of ways in which C1h and C1w also sit together)

Number of arrangements in which at least one couple sits together = 48 + 48 - 24 = 72

Number of arrangements in which neither couple sits together = 120 – 72 = 48

Probability that neither couple sits together = 48/120 = 2/5

There is an alternative solution of case by case evaluation discussed here: http://www.veritasprep.com/blog/2012/01 ... e-couples/
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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2
Let's call the first couple C and c, the second couple K and k, and the single person S.
Let's seat S in different places and figure out the possible ways to have no couples sit together.
If S sits in the first seat, any of the remaining four people could sit next to S.
However, only two people could sit in the next seat:
the two who don't form a couple with the person just seated).

For example, if we have S K so far, C or c must sit in the third seat.
Similarly, we have only one choice for the fourth seat:
the remaining person who does not form a couple with the person in the third seat.
Because we have seated four people already, there is only one choice for the fifth seat;
the number of ways is 4 X 2 X 1 X 1 = 8.
Because of symmetry, there are also 8 ways if S sits in the fifth seat.

Now let's put S in the second seat.
Any of the remaining four could sit in the first seat.
It may appear that any of the remaining three could sit in the third seat, but we have to be careful not to leave a couple for seats four and five.
For example, if we have C S c so far, K and k must sit together, which we don't want.
So there are only two possibilities for the third seat.
As above, there is only one choice each for the fourth and fifth seats.
Therefore, the number of ways is 4 X 2 X 1 X 1 = 8.
Because of symmetry, there are also 8 ways if S sits in the fourth seat.

This brings us to S in the third seat.
Any of the remaining four can sit in the first seat.
Two people could sit in the second seat (again, the two who don't form a couple with the person in the first seat).
Once we get to the fourth seat, there are no restrictions.
We have two choices for the fourth seat and one choice remaining for the fifth seat.
Therefore, the number of ways is 4 X 2 X 2 X 1 = 16.
We have found a total of 8 + 8 + 8 + 8 + 16 = 48 ways to seat the five people with no couples together;

there is an overall total of 5! = 120 ways to seat the five people, so the probability is = 2/5.
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arnaudl wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

A. 1/5
B. 1/4
C. 3/8
D. 2/5
E. 1/2

TL;DR

Method: Using Sets
P(A or B) = P(A) + P(B) - P(A&B)

Total Cases: 5! = 120
P(1st couple together: AA BBS) = 4! * 2 = 48
P (2nd couple together: BB AAS) = 4! * 2 = 48
P (Both couples together: AA BB S) = 3! * 2 * 2 = 24
P (A or B) = 48 + 48 - 24 = 72
Possible Cases = Total Cases - P(A or B) = 120 - 72 = 48

Veritas Prep Official Solution

You can approach this GMAT problem in different ways. One way is a step-by-step case evaluation. Another is to go the reverse way: count all the arrangements in which at least one couple sits together and subtract that from the total arrangements possible. My method of choice is generally the second one. The only catch is that you have to remember to subtract from the total number of arrangements.

What is the total number of arrangements (without any restrictions)? I hope you remember your basic counting principle and will agree that it should be 5! (Five people arranged in five seats). Now, let’s find out the number of favorable cases.

We will discuss both the methods in detail.

Method 1: Step – by – Step Case Evaluation

Let’s say the two couples are {C1h, C1w} and {C2h, C2w} and the single person is S.

Case 1: S takes the first/last chair

If S takes the first chair, any one of the remaining people can take the chair next to him (4 ways). Say, C1h takes this spot. The next place cannot be occupied by C1w but either of C2h and C2w can occupy it (2 ways). Say, C2h occupies it. The fourth chair can be occupied by only one of the remaining 2 people since C2w cannot take it now (1 way). The last chair has only one person remaining for it.

Total number of acceptable arrangements in which S takes the first chair = 4*2*1 = 8

The case would be exactly the same if S took the fifth seat. Think of it this way: The chairs have seat numbers 1-5. Now the numbers have reversed, 1 switched with 5, 2 switched with 4 and 3 is as it is. Now S is sitting on seat number 5 and we have exactly 8 more arrangements possible.

Total number of acceptable arrangements in which S takes the first or fifth chair = 8*2 = 16

Case 2: S takes the second/fourth chair

If S takes the second seat, any of the remaining four people could sit next to S on either side. However, we need to ensure that both people sitting on either side of S are not a couple i.e. C1h, S, C1w should not occupy the first, second and third seats respectively because then C2h and C2w are left and only 2 adjacent seats are vacant. But they cannot take adjacent seats. This means that there are 4 ways in which the first seat can be occupied — i.e., anyone can take it but there are only 2 ways in which the third seat can be occupied since the person taking the third seat must be from the other couple — i.e., if C1h takes the first seat, only C2h or C2w could take the third one. Now we have 2 people and 2 seats leftover. Fourth seat can be occupied in only one way since if C2h takes the third seat, C2w cannot take it i.e. one of the remaining two people cannot take it. Thereafter, one person and one seat are leftover so the fifth seat can be occupied in one way.

Total number of acceptable arrangements in which S takes the second chair = 4*2*1*1 = 8

The case would be exactly the same if S took the fourth seat.

Total number of acceptable arrangements in which S takes the second or fourth chair = 8*2 = 16

Case 3: S takes the middle seat i.e. the third seat

If S takes the third seat, there are two seats on his left and two on his right. We have to ensure that a couple doesn’t sit on one side and the other side would automatically be couple-free. Any one of the four people can occupy the first seat (say C1h takes it). The second seat can be taken by one person from the other couple i.e. by C2h or C2w so it can be occupied in only 2 ways. Now we have two people leftover and two seats. Either one of them could take the fourth seat so it can be occupied in 2 ways. The fifth seat can be occupied in one way.

Total number of acceptable arrangements in which S takes the third chair = 4*2*2*1 = 16

Total number of favorable arrangements = 16 + 16 + 16 = 48

Total number of arrangements = 120

Probability that neither couple sits together = 48/120 = 2/5

Method 2:

The logic I use here is the one we use to solve SETS questions. It needs a little bit of thought but minimum case evaluations.

There are two couples. We don’t want either couple to sit together. Let’s go the reverse way – let’s make at least one of them sit together. We can then subtract this number from the total arrangements to get the number of arrangements in which neither couple sits together.

Would you agree that it is easy to find the number of arrangements in which both couples sit together? It is. We will work on it in a minute. Let’s think ahead now.

How about ‘finding the number of ways in which one couple sits together?’ Sure we can easily find it but it will include those cases in which both couples are sitting together too. But we would have already found the number of ways in which both couples sit together. We just subtract ‘both couples together’ number from ‘one couple together’ number and get the number of arrangements in which ONLY one couple sits together. Think of SETS here.

Let’s do this now.

Number of arrangements in which both couples sit together: Let’s say the two couples are {C1h, C1w} and {C2h, C2w} and the single person is S. There are three groups/individuals. They can be arranged in 3! ways. But in each couple, husband and wife can be arranged in 2 ways (husband and wife can switch places)

Hence, number of arrangements such that both couples are together = 3!*2*2 = 24

Number of arrangements such that C1h and C1w are together: C1 acts as one group. We can arrange 4 people/groups in 4! ways. C1h and C1w can be arranged in 2 ways (husband and wife can switch places).

Number of arrangements in which C1h and C1w are together = 4! * 2 = 48
But this 48 includes the number of arrangements in which C2h and C2w are also sitting together.
Therefore, number of arrangements such that ONLY C1h and C1w sit together = 48 – 24 = 24
Similarly, number of arrangements such that ONLY C2h and C2w sit together = 24

Number of arrangements in which at least one couple sits together = 24 + 24 + 24 = 72

Number of arrangements in which neither couple sits together = 120 – 72 = 48

Probability that neither couple sits together = 48/120 = 2/5

I believe that the second method is much faster and easier. Nevertheless, it’s good to know and understand both.
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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Bunuel Even though once you and Karishma explained the answer to the problem, it seems extremely logical but to come to that inference while giving the test would take some trial and error and approx 3-3.5 mins if we are v well versed with the basics. Can we thus expect to see such questions on the GMAT?
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Re: Two couples and one single person are seated at random in a row of fiv  [#permalink]

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_________________ Re: Two couples and one single person are seated at random in a row of fiv   [#permalink] 18 Jun 2020, 07:58

# Two couples and one single person are seated at random in a row of fiv   