C) What Is the Probability That at Least One of the Wives Ends Up Sitting Next to Her Husband?
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Two couples and one single person are seated at random in a row of fiv [#permalink] 
Updated on: 22 Jan 2022, 09:50
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2 couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sit down together in side by side chairs?
A. one/v
B. 1/4
C. 3/8
D. 2/v
Due east. 1/2
Originally posted past benjiboo on 25 Sep 2009, 13:07.
Terminal edited by Bunuel on 22 Jan 2022, 09:50, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
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Re: Two couples and one single person are seated at random in a row of fiv [#permalink] 21 November 2009, 13:36
Two couples and one unmarried person are seated at random in a row of v chairs. What is the probability that neither of the couples sits together in adjacent chairs?
A. i/5
B. one/4
C. iii/8
D. ii/5
East. 1/2
Allow's find the opposite probability and decrease it from ane.
Opposite event that neither of the couples sits together is event that at leas ane couple sits together. # of arrangements when at leas i couple sits together is sum of arrangements when EXACTLY 2 couples sit down together and EXACTLY i couples sit together.
Couple A: A1, A2
Couple B: B1, B2
Single person: S
EXACTLY ii couples sit together:
Consider each couple as ane unit: {A1A2}{B1B2}{S}, # of arrangement would be: \(iii!*ii!*ii!=24\). 3! # of dissimilar 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: \(iv!*ii!=48\). 4! # of unlike organisation of these 4 units, 2! arrangement of couple A (A1A2 or A2A1). But these 48 arrangements will also include arrangements when 2 couples sit down together, so total for couple A would be \(48-24=24\);
The same for couple B: {B1B2}{A1}{A2}{S}, # of organisation would be: \(4!*two!=48\). Once again these 48 arrangements will too include arrangements when 2 couples sit together, and so full for couple B would be \(48-24=24\);
\(24+24=48\).
Finally nosotros go the # of arrangements when at least one couple sits together is \(24+48=72\).
Total # of arrangements of v people is \(5!=120\), hence probability of an consequence that at leas ane couple sits together would be \(\frac{72}{120}=\frac{iii}{5}\).
So probability of an event that neither of the couples sits together would exist \(i-\frac{3}{5}=\frac{2}{five}\)
Respond: D.
Hope it's clear.
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Re: Two couples and 1 single person are seated at random in a row of fiv [#permalink] 21 Aug 2010, 02:46
I recall I found a better method by using the "glue strategy" where you take 4 instead of 5 people indiating that 2 always sit side by side to each other. (In the end you take 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
Lookout out!
P(A or B) = P(A) + P(B) - P(A and B)
But at present we P(A and B) to often and nosotros have to delete it once
To calculate P(A and B) nosotros glue both couples together and multiply in the stop by 4
3!x4=24
96-24=72
1-(72/120)=two/v
What do yous guys think?
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Re: Two couples and 1 single person are seated at random in a row of fiv [#permalink] 22 October 2009, 01:37
total outcomes v!
outcomes of couple A sit together: two*4!, outcomes of couple B sit together: ii*iv!, only we must deduct the possibility the 2 couples sit together: iii!*iv
Then outcomes of at least one couple sit together= 48+48-24=72
Outcomes of no couple sit down together: five!-72=48
possibility of no couple sit together: 48/five!=2/v
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Re: Two couples and 1 single person are seated at random in a row of fiv [#permalink] twenty October 2011, 13:46
I used the glue method as referenced in Mgmat volume
Full number of rearranging v ppl in 5 chairs = five! = 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 every bit a single person...and so rearranging 3 ppl in iii positions is 3! = 6
we need to business relationship for eight 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 (viii*iii!)/5! = 2/5.
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Re: Ii couples and one unmarried person are seated at random in a row of fiv [#permalink] 20 Oct 2011, 22:10
benjiboo wrote:
Two couples and one single person are seated at random in a row of 5 chairs. What is the probability that neither of the couples sits together in adjacent chairs?
1/5
1/4
3/8
2/five
1/2
I am giving my solution below which needs a trivial bit of idea but minimum case evaluations. The logic I use hither is the 1 we utilize to solve SETS questions. Allow 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? Information technology is. We will work on it in a minute. Allow me recollect ahead now.
How near 'finding the number of ways in which ane couple sits together?' Sure we can detect it but it will include those cases in which both couples are sitting together besides. But we have already found the number of means in which both couples sit together. We just decrease that number from this number and get the number of ways in which Only one couple sits together. Call back of SETS here.
Allow's practice this at present.
Number of arrangements in which both couples sit together:
Say the couples are C1 and C2. I effort and arrange the loner. He can take positions 1, iii and v. He can sit at 3 places. For each one of these positions, the couples tin switch their places, C1 can switch places within themselves and C2 tin can switch places within themselves. And 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 ane grouping. Arrange 4 people/groups in 4! ways. C1 tin 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
Full number of arrangements = five! = 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 = ii/v
(Ideally, you should see that 24 is 1/fifth of 120 so y'all immediately arrive at ii/5)
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Re: Two couples and one single person are seated at random in a row of fiv [#permalink] 26 Jul 2012, 13:06
Total ways to arrange = 5! = 120
Ways in which 1st couple tin can sit together = 4! X two! = 48
Means in which both couple sit together = iii! Ten 2! X 2! = 24
Means in which ONLY 1st couple tin sit together = 48-24 = 24 = Ways in which But second couple sit together
Ways in which neither couple sits together = Total means – Means in which both sit together –Ways in which merely 1st couple sits together – Ways in which 2d couple sits together
= 120-24 –24-24 = 120-72 = 48
Probability = 48/120 = 2/v
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Re: Two couples and 1 single person are seated at random in a row of fiv [#permalink] 19 Aug 2012, 21:54
ok if nosotros presume that each couples every bit a one person total iii person in a row and the probablity of iii person,which couples seat together, sit down on five chairs is 3/5. Only the question is asking neither of couples seat together so the reverse way 1-3/v=2/5
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Re: Ii couples and ane single person are seated at random in a row of fiv [#permalink] 20 Aug 2012, 05:eleven
arnaudl wrote:
Two couples and 1 single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in next chairs?
A. 1/v
B. 1/4
C. 3/8
D. 2/v
Due east. 1/2
I think the best combinatorial solution was given by heyholetsgo. Groovy and brusque!
I didn't encounter a probabilistic approach, so I tried to work out ane. Here it is:
At that place are five chairs, permit'due south visualize them _ _ _ _ _
There is ane single person, let'south denote him/her by S.
Due south tin can choose from 5 chairs where to sit. In that location are some symmetrical placements for S:
i) Due south _ _ _ _ and _ _ _ _ S (0, 1, or 2 couples can be placed together)
two) _ S _ _ _ and _ _ _S_ (0 or 1 couple tin can exist placed together)
And there is the third case:
three) _ _ South _ _ (0, 1, or 2 couples can be placed together).
And so, let's compute for each case the probability that 0 couple sits together.
1) Due south _ _ _ _ or _ _ _ _ South:
\(\frac{ii}{5}\) probability that S sits on either chair one or chair five.
Now kickoff placing people on the remaining 4 chairs and express the corresponding probabilities:
\(\frac{four}{4}\) for the starting time chair in the sequence of four remaining chairs
\(\frac{two}{3}\) for the second chair (cannot be the kickoff person'south mate)
\(\frac{1}{2}\) for the third chair (cannot be the mate of the second person)
\(\frac{1}{i}\) for the last person to exist placed
In conclusion, the probability of no pairs sittings together when S sits on either chair 1 or 5 is given by
\(\frac{2}{5}*\frac{four}{4}*\frac{ii}{3}*\frac{one}{two}*\frac{i}{ane}=\frac{ii}{xv}.\)
2) _ South _ _ _ or _ _ _ Southward _
Again, two possibilities for S, so \(\frac{ii}{5}\) probability for Southward to sit on either chair ii or 4.
At present outset placing people on the remaining four chairs and limited the corresponding probabilities.
Treat simply the _ South _ _ _ case, the other one has the same probabilities:
\(\frac{4}{iv}\) for the first chair from the left
\(\frac{2}{three}\) for the 2d chair (cannot exist the kickoff person's mate considering then the last ii chairs volition be occupied by the other couple)
\(\frac{ane}{two}\) for the third chair (cannot be the mate of the second person)
\(\frac{1}{1}\) for the last person to exist placed
In decision, the probability of no pairs sittings together when S sits on either chair 2 or 4 is given past
\(\frac{2}{5}*\frac{iv}{4}*\frac{2}{3}*\frac{1}{ii}*\frac{i}{1}=\frac{two}{15}.\)
iii) _ _ South _ _
Here now we just accept ane possibility to identify S, then \(\frac{1}{v}\) probability for Southward to sit down on chair 3.
Starting from the left, place people on chairs:
\(\frac{4}{4}\) for the offset chair from the left
\(\frac{2}{3}\) for the second chair (cannot be the first person's mate)
\(\frac{ii}{2}\) for the third chair (we are already bodacious that we separated the two couples)
\(\frac{ane}{one}\) for the final person to exist placed
In decision, the probability of no pairs sittings together when Southward sits on chair 3 is given past
\(\frac{1}{five}*\frac{iv}{4}*\frac{2}{three}*\frac{ii}{2}*\frac{ane}{1}=\frac{2}{15}.\)
In conclusion, the requested probability is \(three*\frac{two}{15}=\frac{2}{v}.\)
Reply D
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Re: Ii couples and ane single person are seated at random in a row of fiv [#permalink] 31 Oct 2012, 08:10
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 side by side chairs?
A. 1/v
B. 1/4
C. 3/8
D. ii/five
E. 1/ii
Instead of doing so many calculations, why non straight upwards detect the probability that no 1 sits together?
Certain, you can practise the venn diagram approach and detect individual circles and decrease the overlap, but this way was a bit easier for me to understand:
_ _ _ _ _ = 5 positions The single person tin sit in three spots (Later we will account for all the permutations past multiply past 3! = 6)
Southward _ _ _ _ (sitting at the stop)
_ Due south _ _ _ (Sitting 1 spot from the end)
_ _ S _ _ (sitting in the center)
Annotation the remaining options (_ _ _ S _ and _ _ _ _ S) are just inverses of the starting time ii.
So permit's look at the first ane: S _ _ _ _
Due south A1 B1 A2 B2
Southward A1 B2 A2 B1
S B1...
Due south B2...
If we choice A1 for that 2nd spot, then the 3rd spot cannot be A2 since A'due south cannot since side by side 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 volition exist the remaining B.
Notation that A and B must alternate - NO Thing where South is located.
This is true for the other ii forms (_ South _ _ _) besides as (_ _ South _ _ ) - A's and B's must alternate.
So dorsum to S _ _ _ _
That second spot S (_) _ _ _ has 4 options to choose from (A1, A2, B1, and B2). Whichever yous 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 cocky-chosen so in that location is no need to calculate those last spots in this case.
S A1 B1 A2 B2
S A1 B2 A2 B1
So hither'southward 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 _ _ _ _ Due south, etc)
So nosotros accept 8 * 3! = viii * vi = 48 possibilities that no couples sit together
Since there are a full of five positions, the total possibilities is 5! = 120
You can also recall of this every bit (5C1 * 4C1 * 3C1 * 2C1 * 1C1) = five! = 120
So divide the 48 possibilities that no couples sit together by the total 120 possibilities and y'all go:
48/120 = 24/60 = 4 / x = 40%
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Re: Ii couples and one unmarried person are seated at random in a row of fiv [#permalink] 26 May 2013, 23:33
Here the word to CATCH is NEITHER.
After the introduction of this word in question , the valid combination consist of all combination where in that location 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 unmarried entity , and so we accept 3 distinct entity which can be arranged in 3! ways and since couples can also exist re-bundled in them selves so total valid arrangement is 3! * two! *2! and total possible arrangement are five!
Probability = 24/120 = 1/5
Probability that Any ONE couple sits together -
now select i couple and accommodate the effective 4 entity in 4!*2! = 48 , But this will contain the cases which have another couple together which nosotros accept already counted in case above and so constructive arrangements = 48-24 = 24
Similarly chose 2d couple and effective organisation will be 24
Probability = (24+24)/120 = ii/v
required probability = 1- (ane/5+two/5)
= two/5
Promise i was able to explicate it!!!!
Delight laurels KUDOS if this was helpful.
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Re: Two couples and ane unmarried person are seated at random in a row of fiv [#permalink] 10 Dec 2013, 06:39
ulas wrote:
ok if nosotros presume 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 5 chairs is 3/5. Just the question is request neither of couples seat together and then the contrary manner 1-iii/5=two/5
Keep It Stupid Simple
Is this approach right? Information technology is like to what I idea merely didn't quite become it right
Cheers
J 
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Re: Ii couples and ane single person are seated at random in a row of fiv [#permalink] x Dec 2013, 08:fifteen
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-three/5=2/5
Keep It Stupid Simple
Is this approach correct? It is similar to what I thought simply didn't quite get it right
Cheers
J
No, that's non correct. The probability that neither of the couples sits together in adjacent chairs does not equal to one- {the probability that exactly two couples sit together}. Information technology'due south i - {the probability that exactly 2 couples sit together} - {the probability that exactly ane couple sits together}.
Also, the probability that EXACTLY 2 couples sit together is 24/5! not 3/5:
Consider each couple as ane unit: {A1A2}{B1B2}{S}, # of arrangement would be: \(3!*2!*2!=24\). 3! # of unlike arrangement of these iii units, two! arrangement of couple A (A1A2 or A2A1), 2! arrangement of couple B (B1B2 or B2B1).
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Re: Two couples and one unmarried person are seated at random in a row of fiv [#permalink] 12 October 2015, 21:14
mohan514 wrote:
Ii 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. i/4
C. iii/eight
D. 2/5
E. 1/2
Responding to a pm:
I am guessing yous are looking for a one line solution since the answer is a elementary 2/five. I couldn't think of the logic that will help united states arrive at the answer direct. I use sets to solve such questions. It'southward non very different from what Bunuel has used higher up.
Hither is my solution:
This question isn't very unlike from our regular "five people sit down in a row such that A does not sit next to B. In how many ways is this possible?"
Hither, there are ii pairs of people who cannot sit next to each other. Hence, yous need to take special care of the cases in which both sit with each other.
In that location are 2 couples. We don't desire either couple to sit together. Permit'southward go the reverse way – allow's make at least one of them sit down together. We can so decrease this number from the total arrangements to go the number of arrangements in which neither couple sits together.
Would you agree that it is easy to notice the number of arrangements in which both couples sit together? It is. We will work on it in a minute. Allow's think ahead for now.
How near 'finding the number of means in which ane couple sits together?' Sure we tin hands find information technology but it will include those cases in which both couples are sitting together too. Simply we would have already found the number of ways in which both couples sit together. When nosotros just subtract 'both couples together' number once from the full to avoid double counting, we will get the number of means in which at to the lowest degree one couple sits together. Think of SETS here.
Let's do this at present.
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 Due south. At that place are iii groups/individuals. They can be arranged in 3! ways. Just in each couple, husband and wife can be bundled in ii ways (husband and wife can switch places)
Hence, number of arrangements such that both couples are together = iii!*2*2 = 24
Number of arrangements such that C1h and C1w are together: C1 acts as ane grouping. We can conform 4 people/groups in four! means. C1h and C1w tin can be arranged in two ways (husband and wife can switch places).
Number of arrangements in which C1h and C1w are together = iv! * 2 = 48
Merely this 48 includes the number of arrangements in which C2h and C2w are also sitting together.
C2h and C2w too sit down together in 48 ways (including the number of ways in which C1h and C1w also sit down together)
Number of arrangements in which at to the lowest degree 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
At that place is an alternative solution of example by example evaluation discussed hither:
http://www.veritasprep.com/web log/2012/01 ... east-couples/
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Re: Two couples and one unmarried person are seated at random in a row of fiv [#permalink] 13 Aug 2017, 02:25
Allow's call the first couple C and c, the second couple K and k, and the single person South.
Let'south seat S in different places and figure out the possible ways to accept no couples sit together.
If South sits in the get-go seat, whatsoever of the remaining four people could sit side by side to S.
Nonetheless, only two people could sit in the next seat:
the two who don't grade a couple with the person only seated).
For case, if nosotros take Southward K and then far, C or c must sit in the 3rd seat.
Similarly, we have but one pick for the quaternary seat:
the remaining person who does non 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 means is 4 10 ii X 1 X one = eight.
Considering of symmetry, in that location are also 8 ways if S sits in the fifth seat.
Now let'due south put S in the 2d seat.
Any of the remaining four could sit in the start seat.
It may appear that any of the remaining three could sit in the third seat, but nosotros have to be careful non to exit a couple for seats four and five.
For example, if we accept C S c so far, K and k must sit together, which we don't want.
So there are just two possibilities for the third seat.
Equally to a higher place, there is only ane pick each for the fourth and fifth seats.
Therefore, the number of means is 4 X two X i 10 1 = viii.
Because of symmetry, there are also 8 ways if S sits in the fourth seat.
This brings the states to S in the third seat.
Any of the remaining four can sit in the offset seat.
Two people could sit in the second seat (again, the ii who don't class a couple with the person in the beginning seat).
One time we become to the quaternary seat, at that place are no restrictions.
We accept two choices for the fourth seat and 1 choice remaining for the fifth seat.
Therefore, the number of means is 4 X 2 Ten ii X i = xvi.
We take found a total of 8 + eight + eight + 8 + xvi = 48 ways to seat the 5 people with no couples together;
there is an overall total of 5! = 120 ways to seat the five people, and then the probability is = 2/5.
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Re: Two couples and one unmarried person are seated at random in a row of fiv [#permalink] 26 May 2019, 12:27
arnaudl wrote:
Two couples and 1 single person are seated at random in a row of 5 chairs. What is the probability that neither of the couples sits together in adjacent chairs?
A. 1/5
B. ane/4
C. 3/eight
D. ii/5
E. 1/ii
TL;DR
Method: Using Sets
P(A or B) = P(A) + P(B) - P(A&B)
Total Cases: v! = 120
P(1st couple together: AA Bulletin board system) = iv! * 2 = 48
P (second couple together: BB AAS) = 4! * 2 = 48
P (Both couples together: AA BB S) = 3! * ii * 2 = 24
P (A or B) = 48 + 48 - 24 = 72
Possible Cases = Total Cases - P(A or B) = 120 - 72 = 48
Answer = 48/120 = 2/v
Veritas Prep Official Solution
Yous tin arroyo this GMAT problem in dissimilar means. One fashion is a step-by-footstep case evaluation. Another is to go the reverse style: 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 i. The merely catch is that you have to call up to subtract from the total number of arrangements.
What is the total number of arrangements (without any restrictions)? I hope you call up your basic counting principle and will agree that information technology should exist five! (Five people arranged in 5 seats). At present, let's find out the number of favorable cases.
Nosotros will discuss both the methods in detail.
Method 1: Pace – past – Pace Case Evaluation
Let's say the ii couples are {C1h, C1w} and {C2h, C2w} and the unmarried person is South.
Case 1: S takes the outset/last chair
If S takes the get-go chair, whatsoever one of the remaining people can have 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 (two ways). Say, C2h occupies it. The fourth chair can be occupied by only 1 of the remaining 2 people since C2w cannot take it now (1 style). The last chair has only one person remaining for information technology.
Total number of acceptable arrangements in which Southward takes the first chair = 4*two*1 = viii
The instance would exist exactly the same if South took the fifth seat. Think of it this way: The chairs have seat numbers one-5. Now the numbers have reversed, 1 switched with five, 2 switched with iv and 3 is as it is. Now Southward is sitting on seat number v and nosotros take exactly viii more arrangements possible.
Total number of adequate arrangements in which S takes the offset or fifth chair = 8*two = 16
Case 2: South takes the second/quaternary chair
If South takes the second seat, any of the remaining four people could sit down side by side to S on either side. Yet, we need to ensure that both people sitting on either side of S are not a couple i.due east. C1h, S, C1w should not occupy the first, second and third seats respectively because and then C2h and C2w are left and only 2 adjacent seats are vacant. But they cannot accept adjacent seats. This means that there are four ways in which the first seat can exist occupied — i.due east., anyone can take it but there are only two ways in which the tertiary seat can exist occupied since the person taking the 3rd seat must be from the other couple — i.e., if C1h takes the outset seat, only C2h or C2w could take the third 1. Now we accept 2 people and 2 seats leftover. Fourth seat tin be occupied in only ane way since if C2h takes the tertiary seat, C2w cannot have it i.east. one of the remaining ii people cannot take information technology. Thereafter, one person and 1 seat are leftover so the 5th seat can be occupied in one way.
Total number of acceptable arrangements in which S takes the second chair = 4*2*1*ane = 8
The instance would be exactly the same if S took the fourth seat.
Full number of acceptable arrangements in which Due south takes the 2d or fourth chair = eight*ii = 16
Case 3: S takes the middle seat i.east. the third seat
If S takes the third seat, there are 2 seats on his left and two on his right. Nosotros have to ensure that a couple doesn't sit on i side and the other side would automatically be couple-gratuitous. Any one of the four people tin can occupy the first seat (say C1h takes it). The 2d seat can exist taken by 1 person from the other couple i.e. by C2h or C2w so it tin can exist occupied in only two ways. At present we have two people leftover and two seats. Either i of them could take the fourth seat so it can be occupied in 2 ways. The 5th seat tin can be occupied in ane style.
Total number of acceptable arrangements in which South takes the third chair = four*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/five
Method 2:
The logic I utilize hither is the one we use to solve SETS questions. It needs a piddling bit of thought but minimum case evaluations.
There are two couples. We don't desire either couple to sit together. Permit'south go the opposite manner – let's brand at to the lowest degree one of them sit together. We can and then decrease this number from the full arrangements to become the number of arrangements in which neither couple sits together.
Would you concur that it is easy to notice the number of arrangements in which both couples sit down together? It is. Nosotros volition piece of work on information technology in a minute. Let's retrieve alee now.
How about 'finding the number of ways in which one couple sits together?' Certain we can hands observe it but it will include those cases in which both couples are sitting together too. Just nosotros would accept already institute the number of ways in which both couples sit together. We merely subtract 'both couples together' number from 'ane couple together' number and get the number of arrangements in which But one couple sits together. Think of SETS here.
Let'southward do this at present.
Number of arrangements in which both couples sit together: Allow's say the two couples are {C1h, C1w} and {C2h, C2w} and the single person is Southward. There are three groups/individuals. They can be arranged in 3! ways. Only in each couple, husband and wife tin can be arranged in 2 ways (husband and wife can switch places)
Hence, number of arrangements such that both couples are together = 3!*ii*2 = 24
Number of arrangements such that C1h and C1w are together: C1 acts every bit one grouping. We can arrange 4 people/groups in 4! ways. C1h and C1w can be bundled in two ways (husband and wife can switch places).
Number of arrangements in which C1h and C1w are together = iv! * 2 = 48
Only this 48 includes the number of arrangements in which C2h and C2w are also sitting together.
Therefore, number of arrangements such that Just 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] 07 Apr 2020, 22:41
Bunuel Fifty-fifty though once you and Karishma explained the answer to the problem, it seems extremely logical but to come up to that inference while giving the test would have some trial and error and approx iii-three.5 mins if we are v well versed with the basics. Tin can we thus look 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] 28 Oct 2020, 20:49
Set Theory
Neither tin be read as (in terms of venn diag) = - A - B + A ∩ B (or A intersection B or A and B tin happen simulatenously). Property used A U B = A + B - A ∩ B + Neither . Here A U B = 5!
A= 1st couple together only
B = second couple together only
....
so A tin exist written as {XY}, P, Q,R = 4!*two!
B can exist written every bit {PQ}, X, Y,R = 4!*2!
and A ∩ B can be written as {XY},{P,Q},R = 3!*2!*ii!
Fav out comes = five ! - 4!*ii! - 4!*ii! + three!*2!*2!
Total outcome = 5!
Ans = [120 - 72] / 120 = 48/120 = viii/20 = 2/five
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Re: Two couples and 1 single person are seated at random in a row of fiv [#permalink] 31 Oct 2020, 03:21
hi VeritasKarishma, I would like to know if the following solution is correct? I used combination in this.
Since there are five seats so the total number of possible outcomes is five! = 120
(both couples sit together)2C1 x 1C1 x 2! 10 2! x3! = 48
(one couple sits together)2C1 10 3 ten 2 x 1 ten two! = 24
( one- couples sit together) ane- 72/120 = 2/5
I originally thought that if i summate the outcomes where the couples sit together then I know the possible outcomes of them not siting together at all.
Thanks!!
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Re: Two couples and ane single person are seated at random in a row of fiv [#permalink] 02 Nov 2020, 22:29
hkkat wrote:
how-do-you-do VeritasKarishma, I would like to know if the following solution is correct? I used combination in this.
Since in that location are 5 seats so the total number of possible outcomes is 5! = 120
(both couples sit together)2C1 x 1C1 x ii! x 2! x3! = 48
(ane couple sits together)2C1 10 iii 10 2 x ane x 2! = 24
( 1- couples sit together) 1- 72/120 = 2/v
I originally thought that if i calculate the outcomes where the couples sit together and then I know the possible outcomes of them not siting together at all.
Thanks!!
Number of mode is which both couples sit together is iii! * two! *2!. You lot have taken an extra 2C1. I am assuming you are selecting a couple. But y'all don't accept to select. They are not getting unlike treatment.
You have C1h and C1w and C2h and C2w. You combine them into 2 units - C1 and C2.
You adapt C1, C2 and South in three! ways.
Arrange C1h and C1w inside C1 in 2 ways and C2h and C2w inside C2 in 2 ways.
No pick needed.
Also when you lot discover the number of cases in which ane couple sits together,
2C1 * iv! * 2!
You select a couple and have 1 pair now. So total four things to be arranged.
Note that it includes the cases where the other couple too sits together.
I have discussed this method in this post:
https://gmatclub.com/forum/two-couples- ... l#p1585549
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Re: Two couples and one single person are seated at random in a row of fiv [#permalink]
02 Nov 2020, 22:29
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