Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 05 Apr 2012
Posts: 39

3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
Updated on: 12 Nov 2014, 09:09
Question Stats:
50% (02:16) correct 50% (02:11) wrong based on 245 sessions
HideShow timer Statistics
3 persons (1 couple and 1 single) are seated at random in a row of 5 chairs. What is the probability that the couple does not sit together? A. 5/7 B. 4/5 C. 2/5 D. 3/5 E. 11/18
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by keiraria on 20 Apr 2012, 16:59.
Last edited by Bunuel on 12 Nov 2014, 09:09, edited 2 times in total.
Edited the question and added answer choices




Math Expert
Joined: 02 Sep 2009
Posts: 58402

Re: 3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
20 Apr 2012, 23:39
keiraria wrote: hello
3 persons  one couple and one single are seated at random in a row of 5 seats
what is the probability that the couple does not sit together
please explain your approach
best regards Let's find the probability that a couple sits together (right next to each other) and subtract that value from 1. Total # of ways 3 persons \(C_1\), \(C_2\) and \(S\) to be seated in a row of 5 seats is \(\frac{5!}{2!}=60\). Consider this, we are interested in arrangement of \(C_1, \ C_2, \ S, \ E, \ E\), so in arrangement of 5 letters out of which 2 E's are identical (E denotes an empty seat); # of ways for a couple to sit together is \(\frac{4!}{2!}*2=24\). Consider a couple as a single unit: \(\{C_1,C_2\}, \ S, \ E, \ E\), so we have total of 4 units out of which 2 E's are identical, # of arrangement of these units is \(\frac{4!}{2!}\), but \(C_1\), \(C_2\) within their unit can be arranged in 2 ways (\(\{C_1,C_2\}\) or \(\{C_2,C_1\}\)), so total # of arrangement for this case is \(\frac{4!}{2!}*2=24\); \(P=1\frac{24}{60}=\frac{3}{5}\). Answer: D.
_________________




Senior Manager
Joined: 13 Aug 2012
Posts: 401
Concentration: Marketing, Finance
GPA: 3.23

Re: 3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
28 Dec 2012, 19:42
keiraria wrote: 3 persons (1 couple and 1 single) are seated at random in a row of 5 chairs. What is the probability that the couple does not sit together?
A. 5/7 B. 4/5 C. 2/5 D. 3/5 E. 11/18 Given: {H} {W} {S} will attempt to seat on _ _ _ _ _ seats How many ways for {H} {W} {S} to seat on _ _ _ _ _ seats? 5*4*3 How many ways for {HW} {S} to seat together on _ _ _ _ _ seats? 4*3 Then multiply by 2 to account for the arrangement of HW. 4*3*2! What is the probability of {H}{W} NOT seating together? \(=1  \frac{4*3*2}{5*4*3} = 1  \frac{2}{5} = 3/5\) Answer: D
_________________
Impossible is nothing to God.



Manager
Joined: 12 Jan 2013
Posts: 56
Location: United States (NY)
GPA: 3.89

Re: 3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
14 Jan 2013, 01:04
The single person is no different from an empty chair Thus, there are \(\frac{{5*4}}{2}=10\) ways to pick two chairs for the couple, but only 4 in which they sit together (CCEEE, ECCEE, EECCE, EEECC). 1  4/10 = 3/5 is the answer.
_________________
Sergey Orshanskiy, Ph.D. I tutor in NYC: http://www.wyzant.com/Tutors/NY/NewYork/7948121/#ref=1RKFOZ



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India

Re: 3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
14 Jan 2013, 01:52
SergeyOrshanskiy wrote: The single person is no different from an empty chair Thus, there are \(\frac{{5*4}}{2}=10\) ways to pick two chairs for the couple, but only 4 in which they sit together (CCEEE, ECCEE, EECCE, EEECC). 1  4/10 = 3/5 is the answer. The perspective we often use to solve such questions is this: The vacant seats are no different from two identical people. Assume that each vacant spot is taken by an imaginary person V. 'The single person is no different from an empty chair' is a refreshing perspective that we can use! Good point! Assume the rest of the three chairs are vacant. Since it is a probability question, the probability we will obtain will be correct. Mind you, the number of ways in which you can arrange a couple and an individual is not given by 10. It is given by 60 only (as shown by Bunuel above). But 10 is the number of ways in which we can choose 2 seats for a couple. In 4 of those 10 ways, the seats will be next to each other and in 6 cases they will not be. Hence the probability obtained will be 3/5. You can also think that you can make the husband and the wife sit in 2 of the 5 chairs in 5*4 = 20 ways. Out of these, in 4*2 = 8 ways, they will sit next to each other. In 12 ways, they will not sit next to each other. So probability will still remain 12/20 = 3/5 As discussed before, in probability questions, whatever logic you use to get the numerator, use the same logic to get the denominator.
_________________
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 >



Manager
Joined: 23 Sep 2015
Posts: 81
Concentration: General Management, Finance
GMAT 1: 680 Q46 V38 GMAT 2: 690 Q47 V38
GPA: 3.5

3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
23 Oct 2015, 22:15
why for this question are we doing 5!/2! for the chairs to people?
I recall from a circular question, when there was 4 men and 4 women sitting around the table, and we had to calculate the number of ways if all the women sat together.
which was (41)! * 4C1 * 4!
So why is this question 5!/2! instead of 5C3 ?
We can incorporate this question into my confusion surrounding this "http://gmatclub.com/forum/therearexpeopleandychairsinaroomwherexandyare170525.html"
There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?
(1) x + y = 12. Since x and y are primes, then x=5 and y=7 OR x=7 and y=5.
If x=5 and y=7, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 chairs from 7, and 5! is the number of arrangements of 5 people on those chairs.
If x=7 and y=5, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 people who will get the chairs, and 5! is the number of arrangements of 5 people on those chairs.



CEO
Joined: 20 Mar 2014
Posts: 2597
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: 3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
25 Oct 2015, 16:14
GMATDemiGod wrote: why for this question are we doing 5!/2! for the chairs to people?
I recall from a circular question, when there was 4 men and 4 women sitting around the table, and we had to calculate the number of ways if all the women sat together.
which was (41)! * 4C1 * 4!
So why is this question 5!/2! instead of 5C3 ?
We can incorporate this question into my confusion surrounding this "http://gmatclub.com/forum/therearexpeopleandychairsinaroomwherexandyare170525.html"
There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?
(1) x + y = 12. Since x and y are primes, then x=5 and y=7 OR x=7 and y=5.
If x=5 and y=7, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 chairs from 7, and 5! is the number of arrangements of 5 people on those chairs.
If x=7 and y=5, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 people who will get the chairs, and 5! is the number of arrangements of 5 people on those chairs. I think you are confusing between straight up arrangements of n things in a row = n! and arrangement of n people around a table = circular arrangement = (n1)! For the question : 3 people (1 couple and 1 single), we are doing 5!/2! to account for 5 'things' to arrange (C1,C2, S,E,E where Es are empty seats) and 2! accounts for the 2 empty seats. You can not do 5C3 as this will remove the cases when you have EEC1C2S or EC1C2ES etc. The other question you are referring to is a bit different. I would suggest that you read the solution ( therearexpeopleandychairsinaroomwherexandyare170525.html#p1358628) carefully and mention your doubt in that thread.



Current Student
Status: DONE!
Joined: 05 Sep 2016
Posts: 357

Re: 3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
07 Dec 2016, 19:55
Total ways to arrange 3 people with 5 total slots = 5C3 = 10
Ways to arrange the couple, so they have a space in between:
(1) X/Y __ Y/X __ __ (2 ways) (2) __ __ X/Y __ Y/X (2 ways) (3) __ X/Y __ Y/X __ (2 ways)
6 total ways to separate the couple
6/10 = 3/5
D.



Intern
Joined: 26 Jun 2012
Posts: 5

Re: 3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
Show Tags
24 Mar 2019, 14:23
I understand how you come to the total amount of ways (60) for the denominator, but how can you acknowledge that the two empty (E) seats are identical when calculating 4!/2! *2, but not acknowledge it when calculating 5!/2! = 60?. Bunuel wrote: keiraria wrote: hello
3 persons  one couple and one single are seated at random in a row of 5 seats
what is the probability that the couple does not sit together
please explain your approach
best regards Let's find the probability that a couple sits together (right next to each other) and subtract that value from 1. Total # of ways 3 persons \(C_1\), \(C_2\) and \(S\) to be seated in a row of 5 seats is \(\frac{5!}{2!}=60\). Consider this, we are interested in arrangement of \(C_1, \ C_2, \ S, \ E, \ E\), so in arrangement of 5 letters out of which 2 E's are identical (E denotes an empty seat); # of ways for a couple to sit together is \(\frac{4!}{2!}*2=24\). Consider a couple as a single unit: \(\{C_1,C_2\}, \ S, \ E, \ E\), so we have total of 4 units out of which 2 E's are identical, # of arrangement of these units is \(\frac{4!}{2!}\), but \(C_1\), \(C_2\) within their unit can be arranged in 2 ways (\(\{C_1,C_2\}\) or \(\{C_2,C_1\}\)), so total # of arrangement for this case is \(\frac{4!}{2!}*2=24\); \(P=1\frac{24}{60}=\frac{3}{5}\). Answer: D.




Re: 3 persons (1 couple and 1 single) are seated at random in a
[#permalink]
24 Mar 2019, 14:23






