Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 05 Jul 2015, 13:25

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Two Couples and One Single Person are seated at random in a

Author Message
TAGS:
Manager
Joined: 25 Aug 2004
Posts: 171
Location: MONTREAL
Followers: 1

Kudos [?]: 11 [0], given: 0

Two Couples and One Single Person are seated at random in a [#permalink]  03 Aug 2005, 05:31
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 adjacents chairs?

(A) 1/5

(B) 2/5

(C) 4/5

(D) 5/7

(E) 3/4
Manager
Joined: 28 Aug 2004
Posts: 205
Followers: 1

Kudos [?]: 1 [0], given: 0

C

ways together:
randomly = 4! = 24 and
a couple next to a couple like in x1x2y1y2... = 16

total together = 8+16 = 24

Pr not together = 1-together = 1 - 24/5! = 4/5.
Senior Manager
Joined: 30 May 2005
Posts: 277
Followers: 1

Kudos [?]: 11 [0], given: 0

Dan wrote:
C

ways together:
randomly = 4! = 24 and
a couple next to a couple like in x1x2y1y2... = 16

total together = 8+16 = 24

Pr not together = 1-together = 1 - 24/5! = 4/5.

plz dan why do you divide 24 by 5! thanks
Senior Manager
Joined: 04 May 2005
Posts: 284
Location: CA, USA
Followers: 1

Kudos [?]: 23 [0], given: 0

A

I felt this is equivalent to ask what is the probability of having
the single sitting in the middle seat. (that is the only situation
where the two couples are not sitting next to each other)

so that is 1 in 5, thus 1/5

the couple together's probability should be quite small,
4/5 seems to be too high.
Manager
Joined: 21 Mar 2005
Posts: 115
Location: Basel
Followers: 1

Kudos [?]: 6 [0], given: 0

A also . Same reason as qpoo.
Manager
Joined: 23 Apr 2005
Posts: 209
Followers: 1

Kudos [?]: 3 [0], given: 0

I go with A.

The total outcomes are 5!

The number of the event that there is a couple sitting adjacently each other is 6*4*2*2=96.

Considering that:

The five people are A,A', B,B',C with AA' are a couple and BB' are the other couple. The five seats are 1,2,3,4,5.

We can see if AA' sitting in two adjacent seats, there are six ways (6) to arrange the remainders in the other three seats.

Moreover, there are four ways (4) to seat A and A' in five seats so that A and A' sit adjacently.

Not only that, there are two ways (2) to seat A and A' adjacently.

Finally, there are two couples (2)

Therefore, the probability we have to calculate is (120-96)/120=1/5

It means that the answer is A
Director
Joined: 05 Jan 2005
Posts: 561
Followers: 2

Kudos [?]: 12 [0], given: 0

Re: Probability- Tough one ! [#permalink]  04 Aug 2005, 04:07
mbassmbass04 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 adjacents chairs?

(A) 1/5

(B) 2/5

(C) 4/5

(D) 5/7

(E) 3/4

Please see http://www.gmatclub.com/phpbb/viewtopic ... 9976#89976 and http://www.gmatclub.com/phpbb/viewtopic ... gle+person for some elegant explainations.
Senior Manager
Joined: 04 May 2005
Posts: 284
Location: CA, USA
Followers: 1

Kudos [?]: 23 [0], given: 0

ah, thanks for the posting Arsene_Wenger

I misread the problem as having two couplese being together.
(couple A) (couple B) single
Senior Manager
Joined: 30 May 2005
Posts: 373
Followers: 1

Kudos [?]: 7 [0], given: 0

To solve this problem we need to find the total # of ways in which at least one couple is seated together and subtract it from the total# of outcomes.

Total # of outcomes = 5! = 120

If A = # of ways couple 1 is seated together and B = #of ways couple 2 is seated together, then we need N(A U B)

N (AUB) = N(A) + N(B) - N(A int B)

N(A) = N(B) = 2*4! = 48
N (A int B) = 2*2*3! = 24

N(AUB) = 48 + 48 - 24 = 72

Favorable outcomes = 120-72 = 48

P = 48/120 = 2/5
Similar topics Replies Last post
Similar
Topics:
Two couples and one single person are seated at random in a 3 29 Jun 2008, 02:17
2 Two couples and one single person are seated at random in a 10 19 Apr 2008, 09:50
Two couples and one single person are seated at random in a 1 27 Oct 2007, 11:35
Two couples and one single person are seated at random in a 6 05 Sep 2007, 21:49
Two couples and one single person are seated at random in a 13 02 Apr 2006, 16:30
Display posts from previous: Sort by
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}