GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Nov 2018, 18:00

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 November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• Free GMAT Strategy Webinar

November 17, 2018

November 17, 2018

07:00 AM PST

09:00 AM PST

Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
• GMATbuster's Weekly GMAT Quant Quiz # 9

November 17, 2018

November 17, 2018

09:00 AM PST

11:00 AM PST

Join the Quiz Saturday November 17th, 9 AM PST. The Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage.

There are 5 married couples and a group of three is to be

Author Message
SVP
Joined: 30 Apr 2008
Posts: 1830
Location: Oklahoma City
Schools: Hard Knocks
There are 5 married couples and a group of three is to be  [#permalink]

Show Tags

19 Jul 2008, 16:53
2
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 6 sessions

HideShow timer Statistics

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html
_________________

------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Director Joined: 10 Sep 2007 Posts: 887 Re: PS:Comb/Perm Married Couples...not swingers ;-) [#permalink] Show Tags 19 Jul 2008, 17:11 4 2 Total ways of selecting 3 people out of 10 = 10C3 = 10*9*8/3*2 = 120 So final answer cannot be greater than 120. In 3 people only 1 couple is possible. So way of selecting 1 couple out of 5 = 5C1 = 5 Way of selecting 1 person out of remaining 8 = 8C1 = 8 Total ways when there will be 1 couple and 1 different person = 8*5 = 40 Total number of ways in which no people will be from same couple = 120 - 40 = 80 SVP Joined: 30 Apr 2008 Posts: 1830 Location: Oklahoma City Schools: Hard Knocks Re: PS:Comb/Perm Married Couples...not swingers ;-) [#permalink] Show Tags 19 Jul 2008, 17:17 3 I like your approach. When I solved it, I did it a bit differently. There are 3 spots Spot #1 can be any of the 10, because no one has been selected so there is no one to avoid. Spot #2 can choose 1 of only 8 people. 9 people left, but we cannot select the spouse of #1 Spot #3 can choose 1 of only 6 people. 8 people left but we can't select the spouse of #1 or spouse of #2. Now, we have 8 x 10 x 6 = 480 total possibilities, but we're not done. This gives us the Permutation of the event, where we have identified order of #1, #2, #3, etc but the order doesn't matter. So we need to divide out the # of times (or permutations) of when we have the same people but in a different order. The total # of perms of a group of 3 is 3x2 = 6 If you remember we had 80 * 6, and we're dividing out 6, this leaves 80. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

GMAT Club Premium Membership - big benefits and savings

Director
Joined: 27 May 2008
Posts: 527
Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

Show Tags

19 Jul 2008, 20:21
2
another method : calculate probablity and multiply it by total number of ways = 120

probablity of selecting first person = 10/10 (can be anyone)
probability of selecting secd person = 8/9 ( there are 9 left, but we can not take spouse of 1)
probability of selecting third person = 6/8 (there are 8 left but we can take only 6)

P = 10/10 * 8/9 * 6/8 = 2/3

fav ways = P * total ways = 2/3 * 120 = 80 Answer
Senior Manager
Joined: 07 Jan 2008
Posts: 267
Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

Show Tags

20 Jul 2008, 07:13
jallenmorris wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

5 couples:
H1W1 H2W2 H3W3 H4W4 H5W5

group of 3 out of 10 people = 10C3 = 10*9*8/6 = 120

now exclude all the possible arrangements when a Husband and a wife can be in a same group.

Possible such groups:
H1W1H2 H1W1W2 ... H5W5H4 H5W5W4[Thre are 5 pairs and each pair can join with 8 other people to make a BAD husband-wife combinatioin]

5*8=40
12-40 = 80
Director
Joined: 01 Jan 2008
Posts: 595
Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

Show Tags

20 Jul 2008, 08:29
2
choose 3 couples out of 5, then there are 2 options for each couple (husband or wife)
C(5,3)*2^3=(5*4/2)*8=10*8 = 80
Manager
Joined: 30 May 2008
Posts: 53
Re: There are 5 married couples and a group of three is to be  [#permalink]

Show Tags

27 Jul 2013, 14:01
jallenmorris wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

what if the questions changed to something like "there are 5 married couples, how many arrangements are there if a husband and wife must stand next to each other?
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1073
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: There are 5 married couples and a group of three is to be  [#permalink]

Show Tags

27 Jul 2013, 14:22
1
catty2004 wrote:
jallenmorris wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

what if the questions changed to something like "there are 5 married couples, how many arrangements are there if a husband and wife must stand next to each other?

Then to find the total number I would consider each couple as a single entity, so we have 5 single units that can be arranged in 5! ways.
$$A,B,C,D,E$$ or $$A,B,C,E,D$$ and so on ...

But within each single unit you can change to position of the husband and the wife $$(H,W)$$ or $$(W,H)$$, so within each couple you have 2 possible arrangements.

$$Tot=5!*2^5$$
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Math Expert
Joined: 02 Sep 2009
Posts: 50613
Re: There are 5 married couples and a group of three is to be  [#permalink]

Show Tags

28 Jul 2013, 04:17
Non-Human User
Joined: 09 Sep 2013
Posts: 8777
Re: There are 5 married couples and a group of three is to be  [#permalink]

Show Tags

01 Nov 2018, 06:59
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: There are 5 married couples and a group of three is to be &nbs [#permalink] 01 Nov 2018, 06:59
Display posts from previous: Sort by