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Two couples and one single person are seated at random in a

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Manager
Status: exam is close ... dont know if i ll hit that number
Joined: 06 Jun 2011
Posts: 128
Location: India
GMAT Date: 10-09-2012
GPA: 3.2
Two couples and one single person are seated at random in a  [#permalink]

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10 Jan 2012, 07:01
1
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Difficulty:

95% (hard)

Question Stats:

48% (02:30) correct 52% (02:46) wrong based on 163 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

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Joined: 02 Sep 2009
Posts: 58117

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12 Jan 2012, 06:36
13
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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

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.

P.S. mohan514 please DO NOT reword the questions and ALWAYS provide answer choices for PS problems (if available). Thanks.
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Re: Two couples and one single person are seated at random in a  [#permalink]

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12 Oct 2015, 22:14
2
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  [#permalink]

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13 Oct 2015, 05:07
VeritasPrepKarishma wrote:
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/

Thanks Karishma for taking the efforts to explain this. +1
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Joined: 02 Jul 2015
Posts: 102
Schools: ISB '18
GMAT 1: 680 Q49 V33
Re: Two couples and one single person are seated at random in a  [#permalink]

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13 Oct 2015, 07:23
Bunuel 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

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.

P.S. mohan514 please DO NOT reword the questions and ALWAYS provide answer choices for PS problems (if available). Thanks.

Engr2012

Can you help me out with this?

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).

Should this not be 3!*2!*2!*2!?
3! ways for {a1, a2} to take places, 2! ways for them to arrange themselves (a1a2, a2a1), then 2! ways for {b1,b2} to take places and finally 2!ways for them to arrange themselves (b1b2, b2b1)
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Re: Two couples and one single person are seated at random in a  [#permalink]

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13 Oct 2015, 07:42
longfellow wrote:
Bunuel 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

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.

P.S. mohan514 please DO NOT reword the questions and ALWAYS provide answer choices for PS problems (if available). Thanks.

Engr2012

Can you help me out with this?

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).

Should this not be 3!*2!*2!*2!?
3! ways for {a1, a2} to take places, 2! ways for them to arrange themselves (a1a2, a2a1), then 2! ways for {b1,b2} to take places and finally 2!ways for them to arrange themselves (b1b2, b2b1)

You are over counting. For arrangment such as {A1A2}{B1B2}{S}, let {A1A2}=A, {B1B2}=B, then 3! signifies arrangement of ABS (ABS, ASB,BAS,BSA,SAB,SBA) while 2! each will represent internal arrangement of A1A2 and B1B2, giving you a total of 3!*2!*2! = 24 arrangements.

The text in red above is wrong. How can you get 3! =6 ways to arrange 2 entities A1A2?

The last 2! ways as mentioned by you will be taken care of by the 2! that you assume for B1B2.

Hope this helps.
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Two couples and one single person are seated at random in a  [#permalink]

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13 Oct 2015, 16:37
Let A be any person from Couple A and B be any person from Couple B.

There are only two cases how the single person could be distributed to split the couples:

(1) * A1 * B1 * A2 * B2 *
(2) A1 B1 * A2 B2

where * denotes the possible seating arrangements for the single person.

(1) There are $$5 * 2! * 2! * 2!=40$$ different seating arrangements. We multiply by 2! because there are two options for couple A and two options for couple B to be seated. Furthermore, couple B could be seated to the left of A. Consider these possibilities:

* A2 * B1 * A1 * B2 *
* A1 * B2 * A3 * B1 *
* B1 * A1 * B2 * A2 *
etc.

(2) There are $$2! * 2! * 2!=8$$ different seating arrangements. Same reason as above. Consider these possibilities:

A2 B1 * A1 B2
B1 A1 * B2 A2
etc.

Overall we have $$40 + 8 = 48$$ possible arrangements and $$5!=120$$ ways to arrange 5 people. The probability is $$\frac{48}{120}=\frac{2}{5}$$.
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Re: Two couples and one single person are seated at random in a  [#permalink]

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16 Jul 2019, 00:33
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Re: Two couples and one single person are seated at random in a   [#permalink] 16 Jul 2019, 00:33
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