November 22, 2018 November 22, 2018 10:00 PM PST 11:00 PM PST Mark your calendars  All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA) November 23, 2018 November 23, 2018 10:00 PM PST 11:00 PM PST Practice the one most important Quant section  Integer properties, and rapidly improve your skills.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 24 Sep 2009
Posts: 22

Two couples and one single person are seated at random in a
[#permalink]
Show Tags
Updated on: 12 Jul 2013, 12:23
Question Stats:
46% (02:20) correct 54% (01:43) wrong based on 71 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 Explanation: 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 × 2 × 1 × 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 × 2 × 1 × 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 × 2 × 2 × 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 >>>>> 48/120 or 2/5 <<<<<
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by benjiboo on 25 Sep 2009, 13:07.
Last edited by Bunuel on 12 Jul 2013, 12:23, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




Manager
Joined: 11 Aug 2008
Posts: 126

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
22 Oct 2009, 01:37
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+4824=72 Outcomes of no couple sit together: 5!72=48 possibility of no couple sit together: 48/5!=2/5




GMAT Tutor
Joined: 24 Jun 2008
Posts: 1327

Two couples and one single person are seated at random in a
[#permalink]
Show Tags
25 Sep 2009, 15:45
I posted a solution here, which is essentially the same type of case analysis as in the above. There's another solution in the thread below that reverses the problem: count all the arrangements possible, then subtract all the arrangements in which at least one couple does sit together. I don't see a tensecond solution, however.
_________________
GMAT Tutor in Toronto
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com



Manager
Joined: 27 Oct 2008
Posts: 179

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
25 Sep 2009, 22:03
I too solved the prob the other way round. The total number of ways that none sit together = 5!  (Total number of ways in which atleast one pair sits together) = 5!  72 = 48
thus the probability is = 48/120 = 2/5



Manager
Joined: 11 Sep 2009
Posts: 129

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
25 Sep 2009, 23:12
I got 2/5 as well, but I did it in a slightly different way. Maybe this approach is easier for some people, although I find Ian's explanation to be quite clear as well.
Basically, you have 1 single person, and two couples (couple A and B). So 5 people: A A B B and S. You want no repeating characters beside each other.
They sit in 5 chairs: _ _ _ _ _
3 Cases (based on where the single person is located)
Case 1: S in the middle (P = 1/5)
_ _ [S] _ _ [A or B] _ [S] _ _ [A or B] [2/3 (opposite couple)] [S] [Doesn't Matter] [Doesn't Matter]
P = 2/3
Case 2: S on the End (P = 2/5)
[S] _ _ _ _ [S] [A or B] _ _ _ [S] [A or B] [2/3 (opposite couple)] [1/2 (opposite couple)] [Doesn't Matter]
P = 2/3 * 1/2 = P = 1/3
Case 3: S Next to the End (P = 2/5)
_ _ _ [S] _ [A or B] _ _ [S] _ [A or B] [2/3 (opposite couple)] [1/2 (opposite couple)] [S] [Doesn't Matter]
P = 2/3 * 1/2 = P = 1/3
Total Probability = (1/5)*(2/3) + (2/5)*(1/3) + (2/5)*(1/3) = (2/15) + (2/15) + (2/15) = 2/5



VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1040

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
09 May 2011, 00:51
2! * 4! + 2!*4!  3!*2! * 2! = 72 12072= 48 48/5!



Intern
Joined: 20 Oct 2011
Posts: 1

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
20 Oct 2011, 12:48
I have another way with just 2 cases to consider (with the same final answer of 2/5): For simplicity I'll rephrase the question as putting 2 A's, 2 B's, and 1 C in a row such that two A's are not adjacent and the two B's are not adjacent. _ _ _ _ _ 1 2 3 4 5
case 1: The letters on either end of the row (positions 1 and 5) are not the same: This is equal to having no adjacent letters the same in the following diagrams:
_ _ _ _ _ 1 2 3 4 5
_ _ _ _ _ 2 3 4 5 1
_ _ _ _ _ 3 4 5 1 2
_ _ _ _ _ 4 5 1 2 3
_ _ _ _ _ 5 1 2 3 4
I.e. the pairs 1 and 2, 2 and 3, 3 and 4, 4 and 5, 5 and 1 can not be the same.
If we place C on one of the 5 positions (this is 5 choices) we use the diagram which would have C in the middle. Now, we have only two possibilities for the position right of C , A or B:
_ _ C ? _
If ? is A, the only possibility is
A B C A B
likewise, if ? is B, the only possibility is
B A C B A
therefore there are only 2 total possibilities. 5 (places from c)*2 (possibilites per position of c) =10 total possibilities for the case.
case 2: The letters on positions 1 and 5 are the same: _ _ _ _ _ 1 2 3 4 5
The only 2 possibilites are
A B C B A
or
B A C A B
so 2 total possibilities for this case.
In conclusion:
10+2=12 total favourable possibilities
the number of ways to place 2A's, 2B's and a C is
5 (places for the C)*(4*3/2)(ways to place the two A's)=30. Note that after we placed the C and the A's, the B's have only a single way to be placed.
so, 12/30 = 2/5.



Manager
Status: MBAing!!!!
Joined: 24 Jun 2011
Posts: 219
Location: United States (FL)
Concentration: Finance, Real Estate
Schools: Wharton '14 (D), CBS '14 (WL), Ross '14 (WL), Haas '14 (D), Johnson '14 (A), McCombs '14 (II), KenanFlagler '14 (M), Madison (A), Consortium (D), Consortium (A), Consortium (M), Consortium (II), Consortium (A), Consortium (WL)
GPA: 3.65
WE: Project Management (Real Estate)

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
20 Oct 2011, 13:46
I used the glue method as referenced in Mgmat bookTotal 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.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8574
Location: Pune, India

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
20 Oct 2011, 22:10
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)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 19 Apr 2010
Posts: 177
Schools: ISB, HEC, Said

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
21 Oct 2011, 22:27
Excellent approach Karishma. At what level in quants one can expect such a question in actual GMAT.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8574
Location: Pune, India

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
27 Oct 2011, 20:11
prashantbacchewar wrote: Excellent approach Karishma. At what level in quants one can expect such a question in actual GMAT. It is definitely a tough one. I would say closer to 750. The reason is that it can be time consuming and confusing if you do step by step case evaluations.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 25 May 2011
Posts: 121

Re: very hard probability / combination (700+) helpppp
[#permalink]
Show Tags
17 Dec 2011, 21:26
Here is my approach:
P=Probability of neither sit together P1=probability of first couple sit together P2=probability of second couple sit together P3=probability of both couples sit together
\(P=1(P1+P2P3)\)
\(P1=P2=\frac{4!*2}{5!}=\frac{2}{5}\)
\(P3=\frac{3!*2*2}{5!}=\frac{1}{5}\)
\(P=1(\frac{2}{5}+\frac{2}{5}\frac{1}{5})= 1\frac{3}{5}= \frac{2}{5}\)



Intern
Joined: 02 Jun 2018
Posts: 6

Re: Two couples and one single person are seated at random in a
[#permalink]
Show Tags
15 Nov 2018, 10:53
ngoctraiden1905 wrote: 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+4824=72 Outcomes of no couple sit together: 5!72=48 possibility of no couple sit together: 48/5!=2/5 Can you please explain how you got 3!*4 ?




Re: Two couples and one single person are seated at random in a &nbs
[#permalink]
15 Nov 2018, 10:53






