GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 21 Feb 2020, 10:23 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Two couples and one single person are seated at random in a

Author Message
TAGS:

### Hide Tags

Intern  Joined: 30 Sep 2009
Posts: 12
Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

8
106 00:00

Difficulty:   95% (hard)

Question Stats: 41% (02:40) correct 59% (02:31) wrong based on 549 sessions

### HideShow timer Statistics

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
Math Expert V
Joined: 02 Sep 2009
Posts: 61385

### Show Tags

37
1
33
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 correct
E: 1/2

Ignoring the constraint first : total number of arrangements = 5! = 120
Then, I try to consider the nb of arrangment where at least one couple sits together but I cannot find using formulas the 72 required. (ie the 3/5 required which leads to the correct answer choice 1 - 3/5 = 2/5) ?

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.
_________________
Manager  Joined: 06 Jul 2010
Posts: 69
Re: 2 Couples and a Single person Probability question  [#permalink]

### Show Tags

7
3
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
Manager  Joined: 26 Oct 2010
Posts: 56

### Show Tags

Two couples and on 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 seem to be able to come up with the right numbers but I can't figure why I would put them together.

5 seats so, 5! options for people to be arranged = 120. And If the one of the couples sit together then there are 4! ways for the others to sit (24), since there are 2 couples then 2x4! (48).

But even if this is right I don't know why I would end up setting the answer so that 4!+4!/5! (48/120) as the probability. How can one convert the different combinations of something into the probability that something happens?

Maybe its just coincidence that that ends up being the right answer.

thanks
Manager  Status: swimming against the current
Joined: 24 Jul 2009
Posts: 133
Location: Chennai, India

### Show Tags

jonblazon wrote:

Two couples and on 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 seem to be able to come up with the right numbers but I can't figure why I would put them together.

5 seats so, 5! options for people to be arranged = 120. And If the one of the couples sit together then there are 4! ways for the others to sit (24), since there are 2 couples then 2x4! (48).

But even if this is right I don't know why I would end up setting the answer so that 4!+4!/5! (48/120) as the probability. How can one convert the different combinations of something into the probability that something happens?

Maybe its just coincidence that that ends up being the right answer.

thanks

probability = chances that incident occurs/total possibilities

Q asks none of them sits together = 1-chances of one sitting together-chances of both sitting together

i.e., $$1-(24/120)-(48/120)$$
Manager  Joined: 02 Oct 2010
Posts: 74

### Show Tags

mailnavin1 wrote:
jonblazon wrote:

Two couples and on 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 seem to be able to come up with the right numbers but I can't figure why I would put them together.

5 seats so, 5! options for people to be arranged = 120. And If the one of the couples sit together then there are 4! ways for the others to sit (24), since there are 2 couples then 2x4! (48).

But even if this is right I don't know why I would end up setting the answer so that 4!+4!/5! (48/120) as the probability. How can one convert the different combinations of something into the probability that something happens?

Maybe its just coincidence that that ends up being the right answer.

thanks

probability = chances that incident occurs/total possibilities

Q asks none of them sits together = 1-chances of one sitting together-chances of both sitting together

i.e., $$1-(24/120)-(48/120)$$

Can you pls explain again..
I was not able to understand the second statement of the explanation...
Math Expert V
Joined: 02 Sep 2009
Posts: 61385

### Show Tags

1
jullysabat wrote:
mailnavin1 wrote:
jonblazon wrote:

Two couples and on 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 seem to be able to come up with the right numbers but I can't figure why I would put them together.

5 seats so, 5! options for people to be arranged = 120. And If the one of the couples sit together then there are 4! ways for the others to sit (24), since there are 2 couples then 2x4! (48).

But even if this is right I don't know why I would end up setting the answer so that 4!+4!/5! (48/120) as the probability. How can one convert the different combinations of something into the probability that something happens?

Maybe its just coincidence that that ends up being the right answer.

thanks

probability = chances that incident occurs/total possibilities

Q asks none of them sits together = 1-chances of one sitting together-chances of both sitting together

i.e., $$1-(24/120)-(48/120)$$

Can you pls explain again..
I was not able to understand the second statement of the explanation...

1-(the probability of the event when EXACTLY 1 couple sits together)-(the probability of the event when EXACTLY 2 couples sit together)=1-48/120-24/120=48/120=2/5.

Complete solution: 2-couples-and-a-single-person-probability-question-92400.html#p654993
_________________
Intern  Joined: 23 Aug 2010
Posts: 10
Re: 2 Couples and a Single person Probability question  [#permalink]

### Show Tags

Thanks to heyholetsgo. I think that P(A or B) = P(A) + P(B) - P(A and B) is the best approach:

Combinations for one couple sitting next to each other 4! x 2! ( couple arrange in 2! ways between themselves)=24x2=48
Combinations for the other couple sitting next to each other 4! x 2! ( couple arrange in 2! ways between themselves)=24x2=48

To calculate P(A and B) we glue both couples together
3! x 2! x 2! ( each couple arrange in 2! ways between themselves)
= 24

P(A) + P(B) - P(A and B)
48 + 48 -24=72
1-(72/120)=2/5
Manager  Joined: 25 Aug 2011
Posts: 134
Location: India
GMAT 1: 730 Q49 V40
WE: Operations (Insurance)
Re: 2 Couples and a Single person Probability question  [#permalink]

### Show Tags

is there something fundamental i am missing out while calculating the probability that exactly 1 person sits together?

if couples are C1,C2 and individual is I..
Now, if only 1 couple has to sit together that couple can be chosen as C1 or C2 therefore 2 ways
so now assuming the couple sitting together as 1 unit. we have 4 people to be arranged in 4 seats so 4P4 ways = 24.
However the couple sitting together may also swap places internally in 2 ways therefore total number of ways = 24X2 = 48 ways and ccounting for the choice of either C1 or C2 sitting together we have 96 ways

now, this includes scenarios in which both couples are sitting together.. therefore for either one to be sitting together I subtract 24. (ways in which both ppl can sit together) to get 72

since total number of ways without any restriction is 120 . probability of at least 1 sitting together is 72/120 = 3/5. so probability of none sitting together is 2/5... Where am i going wrong
Intern  Joined: 13 May 2012
Posts: 25
Schools: LBS '16 (A)
GMAT 1: 760 Q50 V41
Re: 2 Couples and a Single person Probability question  [#permalink]

### Show Tags

1
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
Intern  Joined: 11 Apr 2012
Posts: 2
Location: United States
Concentration: Finance, Entrepreneurship
GMAT Date: 08-11-2012
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

3
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
Director  Joined: 22 Mar 2011
Posts: 583
WE: Science (Education)
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

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 $$3*\frac{2}{15}=\frac{2}{5}.$$

_________________
PhD in Applied Mathematics
Love GMAT Quant questions and running.
SVP  G
Joined: 14 Apr 2009
Posts: 2271
Location: New York, NY
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

2
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%
Manager  Joined: 14 Nov 2011
Posts: 114
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

gmatpill wrote:
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%

Hi, Can you please explain which one of the 6 cases is missing here,

RB1A1B2A2 - 1(R)*2(B's relative to A)*2(arrangement of 2 B's)*2(arrangement of 2 A's) - 8 cases
B1RA1B2A2 - 8
B1A1RB2A2 - 8
B1A1B2RA2 - 8
B1A1B2A2R - 8

I am only able to get 8*5 = 40 cases.

Which one is missing?
Intern  Joined: 13 Feb 2013
Posts: 19
Location: United States
Concentration: Finance, General Management
Schools: HBS '16
GMAT Date: 07-28-2014
GPA: 3.75
WE: Consulting (Computer Software)
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

3
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!!!!

SVP  Joined: 06 Sep 2013
Posts: 1514
Concentration: Finance
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

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 Math Expert V
Joined: 02 Sep 2009
Posts: 61385
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

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).
_________________
Intern  Joined: 19 Mar 2013
Posts: 20
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

My approach
_ _ _ _ _ 5 seats
1 - unfavorable outcomes/ all combinations = favorable outcomes

unfavorable outcomes:
1) one couple sit together 4! 2! = 48
3) both couples sit together 3! 2!2!= 24

48+24=72

1 - 72/120 = 2/5
Intern  Joined: 16 Jan 2016
Posts: 19
Location: United States (CA)
Leonid: B
Concentration: Operations, General Management
GPA: 3.6
WE: Operations (Other)
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

Bunuel wrote:
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 correct
E: 1/2

Ignoring the constraint first : total number of arrangements = 5! = 120
Then, I try to consider the nb of arrangment where at least one couple sits together but I cannot find using formulas the 72 required. (ie the 3/5 required which leads to the correct answer choice 1 - 3/5 = 2/5) ?

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.

How did you get to be so good at Math?
_________________
I Write...

http://www.misterethoughts.blogspot.com
Intern  Joined: 04 Nov 2015
Posts: 12
Re: Two couples and one single person are seated at random in a  [#permalink]

### Show Tags

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

Hi All

Can't we solve this problem like this: (A1,A2,B1,B2,C)

Since no one sits together and we have 5 chairs, i.e. we can arrange people like: _ X _ X _ (where X can be anyone person from the 2 couples)
No of ways of selecting who sits on 2nd chair: 4C1 = 4
No of ways of selecting who sits on 4th chair: 1 (it has to be the other person from couple)
Rest 3 places can be filled in 3! ways = 6
Total ways of arranging 5 people = 5! = 120

Please let me know if my approach is wrong.

Thanks Re: Two couples and one single person are seated at random in a   [#permalink] 29 Aug 2016, 05:19

Go to page    1   2    Next  [ 28 posts ]

Display posts from previous: Sort by

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