Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 22 Jul 2008
Posts: 95
Location: Bangalore,Karnataka

If a committee of 3 people is to be selected from among 5 [#permalink]
Show Tags
05 Jan 2010, 06:46
5
This post received KUDOS
46
This post was BOOKMARKED
Question Stats:
72% (00:53) correct 28% (01:12) wrong based on 1037 sessions
HideShow timer Statistics
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible? A. 20 B. 40 C. 50 D. 80 E. 120
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Aug 2009
Posts: 5660

Re: committee of 3 [#permalink]
Show Tags
05 Jan 2010, 07:06
5
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
ANS 80.. total people=10.. ways to select 3 out of them=10c3=120... it includes comb including couple.. ways in which couple are included = 8c1*5=40.. so ans reqd 12040=80... ( if we take a gp to include a couple ,it will include couple +any one of rest 8 so 8c1 ways .. 5 couple so 5*8c1=40)
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
BANGALORE/



Manager
Joined: 09 May 2009
Posts: 203

Re: committee of 3 [#permalink]
Show Tags
05 Jan 2010, 08:03
15
This post received KUDOS
6
This post was BOOKMARKED
kirankp wrote: If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible? A. 20 B. 40 C. 50 D. 80 E. 120 total no for selecting 3 out of 10=10c3=120 no. of ways in which no two married people included= tot 2 married couple included 2 married couple can be included in 5c1( no. of ways selecting a couple) * 8c1( no. of ways selecting the third person)=5 * 8=40 reqd comb=12040=80 hence D
_________________
GMAT is not a game for losers , and the moment u decide to appear for it u are no more a loser........ITS A BRAIN GAME



Math Expert
Joined: 02 Sep 2009
Posts: 43894

Re: committee of 3 [#permalink]
Show Tags
05 Jan 2010, 09:07
26
This post received KUDOS
Expert's post
31
This post was BOOKMARKED



Manager
Joined: 27 Apr 2008
Posts: 191

Re: committee of 3 [#permalink]
Show Tags
06 Jan 2010, 10:34
9
This post received KUDOS
6
This post was BOOKMARKED
I like to think of it like this:
Step 1  find the combinations without any restrictions
10C3 = 120
Step 2  subtract the combinations that would have a couple in the committee
5C1 x 4C1 x 2 = 40
In this step, we first find the # of ways to choose a couple, which is 5C1=5. After getting the first couple, we need 1 more member, so we choose 1 couple of the remainin 4 couples, which is 4C1 = 4. But within this new couple, we can either choose the man or the woman, so we need to x2.
Step 3  find answer (no restrictions minus restrictions)
120  40 = 80
So the answer is 80.



Manager
Joined: 27 Apr 2008
Posts: 191

Re: committee of 3 [#permalink]
Show Tags
06 Jan 2010, 10:37
Bunuel wrote: If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
Another way to think about this problem:
Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.
But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.
Total # of ways: 5C3*2^3=80. I actually like this way of thinking more though.



Intern
Joined: 20 Dec 2009
Posts: 14

Re: committee of 3 [#permalink]
Show Tags
07 Jan 2010, 04:30
1
This post received KUDOS
1
This post was BOOKMARKED
I too got 80 with the conventional way of 10C3  5C1 * 8C1 = 120  40 = 80. But learnt and loved Bunuel's way. Thanks!



Senior Manager
Joined: 19 Nov 2007
Posts: 457

Re: committee of 3 [#permalink]
Show Tags
27 Feb 2010, 09:39
I understand the 1x approach, but if I were to do it the straighforward way, I get 10 x 8 x 6 (first place 10 ways, second place 8 ways, third place 6 ways) = 480, which is wrong. What am I missing here?
_________________
Underline your question. It takes only a few seconds! Search before you post.



Manager
Joined: 23 Apr 2009
Posts: 67
Location: Texas

Re: committee of 3 [#permalink]
Show Tags
28 Sep 2010, 07:38
Bunuel wrote: If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
Another way to think about this problem:
Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.
But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.
Total # of ways: 5C3*2^3=80. I like this way of thinking and the calculations seem simpler and quicker.



Intern
Joined: 10 Oct 2010
Posts: 23
Location: Texas

Re: committee of 3 [#permalink]
Show Tags
11 Oct 2010, 01:43
5
This post received KUDOS
2
This post was BOOKMARKED
"If a committee of 3 people is to be selected" Combo box arrangement (_)(_)(_)/3!
"from among 5 married couples" Bag of 10 choices: A,B,C,D,E,F,G,H,I,J
"so that the committee does not include two people who are married to each other" First slot has 10 choices (10)(_)(_)/3!
but the choice eliminates the spouse. The second slot has 8 choices (10)(8)(_)/3!
but the choice eliminates another spouse. The third slot has 6 choices (10)(8)(6)/3!
"how many such committees are possible?" (10)(8)(6)/(3*2) = 80



Manager
Joined: 01 Apr 2010
Posts: 164

Re: committee of 3 [#permalink]
Show Tags
15 Jan 2011, 03:28
can you please explain the combo box arrangement explanation for the problem ??
i am not able to understand how we get 3! in the denominator ??



Math Expert
Joined: 02 Sep 2009
Posts: 43894

Re: committee of 3 [#permalink]
Show Tags
15 Jan 2011, 13:55



Manager
Joined: 09 Nov 2012
Posts: 66

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
Show Tags
25 Oct 2013, 18:01
6
This post received KUDOS
Using slot method: First person can be chosen > 10 ways, 2nd person can be chosen > 8 ways (1st person and his wife are not candidates) and 3rd person can be chosen > 6 ways (1st person/2nd person and their wives are out) Answer > 10*8*6/6 = 80 (divide by 6 because the 3 people can be chosen in any order (i.e. 3! = 3*2*1 ways))



Intern
Status: preparing for the GMAT
Joined: 16 Jul 2013
Posts: 37
Concentration: Technology, Entrepreneurship
GMAT Date: 10152013
GPA: 3.53

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
Show Tags
25 Nov 2013, 23:53
I solved the question in different way. first I computed the number of ways of selecting 3 out of 10, which is 120 second I computed the probability of selecting 3 unmarried people out of 5 couples = 10/10 * 8/9 * 6/8 = 2/3 finally multiplying the total number of selection by the probability of selecting 3 unmarried people 2/3 * 120 = 80
_________________
لا الله الا الله, محمد رسول الله
You never fail until you stop trying ,,,



Intern
Joined: 19 Mar 2013
Posts: 22

Re: committee of 3 [#permalink]
Show Tags
12 Dec 2013, 21:29
1
This post received KUDOS
Bunuel, please correct me if i'm wrong. Thank you for your help, i appreciate it!
10*8*6=480 (we chose 3 people out of 10 so that no couple included) 480/3!= 80 (unarranged the order as it doesn't matter)



Intern
Joined: 26 Mar 2014
Posts: 1

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
Show Tags
24 Apr 2014, 12:48
saintforlife wrote: Using slot method: First person can be chosen > 10 ways, 2nd person can be chosen > 8 ways (1st person and his wife are not candidates) and 3rd person can be chosen > 6 ways (1st person/2nd person and their wives are out) Answer > 10*8*6/6 = 80 (divide by 6 because the 3 people can be chosen in any order (i.e. 3! = 3*2*1 ways)) I got this slot method but unfortunately I am unable to get the same answer using the box method. I first find the number of ways I can find a couple: 10(Can choose any person) * 1(Needs to be the spouse of the person chosen in the first place) * 8(Can be any of the remaining 8) = 80/3! Total possible combinations = 10C3 = 120 Therefore number of combinations for unmarried couples = 120(80/3!) which is obviously the wrong answer. Why is this approach wrong ?



Intern
Joined: 20 May 2014
Posts: 37
Location: India

Re: Combinatorics question [#permalink]
Show Tags
22 May 2014, 21:26
2
This post received KUDOS
1
This post was BOOKMARKED
Hi achakrav2694, Ordering is not Required in Selection. We can have just 4 Cases:1) Selecting All 3 husbands: This would be\(5C3 = 10\) 2) Selecting 2 husbands and 1 Wife: \(5C2 * 3\) (As Wife cannot be for the 2 husbands selected) \(= 30\) 3) Selecting All 3Wives: This would be \(5C3 = 10\) 4) Selecting 2 Wives and 1 Husband: \(5C2 * 3\)(As Husband cannot be for the 2 Wives selected) \(= 30\) Total Commitees \(= 10 + 30 + 10 + 30 = 80\) Rgds, Rajat
_________________
If you liked the post, please press the'Kudos' button on the left



Director
Status: Verbal Forum Moderator
Joined: 17 Apr 2013
Posts: 589
Location: India
GMAT 1: 710 Q50 V36 GMAT 2: 750 Q51 V41 GMAT 3: 790 Q51 V49
GPA: 3.3

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
Show Tags
15 Jul 2014, 03:06
1
This post received KUDOS
1
This post was BOOKMARKED
Bunuel wrote: If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
A. 20 B. 40 C. 50 D. 80 E. 120
Another way to think about this problem:
Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) to send only one "representatives" to the committee: 5C3=10.
But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.
Total # of ways: 5C3*2^3=80.
Answer: D. wow superb approach I would give you 100 Kudos. THANKS Bunuel.
_________________
Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/howtoscore750and750imovedfrom710to189016.html



Current Student
Joined: 08 Feb 2014
Posts: 205
Location: United States
Concentration: Finance
WE: Analyst (Commercial Banking)

If a committee of 3 people is to be selected from among 5 [#permalink]
Show Tags
15 Sep 2014, 17:02
1
This post received KUDOS
For those as challenged with math as myself, try this (apologies if this has already been done):
Consider the "slot method"
We need to select three people out a total of ten people (5 couples=10ppl), so create three "slots"
_ _ _ 1 2 3
The only restrictions we have are that we cant pick an individual and that individual's spouse, nor can we select any individual twice. Knowing this, we can put any of our 10 people in the first slot. The second spot, however, will be limited to 8ppl (10 total less the person we placed in spot one, leaves 9, less that person's spouse, leaves 8). The third slot will only have 6 possibilities (10 total less the two ppl already placed and each of their spouses). Fill in the slots:
10 8 6 1 2 3
We would then multiply across to get 10*8*6=480
When using the Slot method, if the order of the selections does not matter (as is the case here) we must divide the the product of the slots by n!, where n is just the number of slots.
480/3! =480/6 = 80
Generally, you'd want to simplify before dividing, so the 2*3 in the 3! term would cancel the 6 in the numerator, and you'd just be left with (10*8)/1 or 80.



Intern
Status: preparing
Joined: 30 Dec 2013
Posts: 41
Location: United Arab Emirates
Concentration: Technology, Entrepreneurship
GMAT 1: 660 Q45 V35 GMAT 2: 640 Q49 V28 GMAT 3: 640 Q49 V28 GMAT 4: 640 Q49 V28 GMAT 5: 640 Q49 V28
GPA: 2.84
WE: General Management (Consumer Products)

Re: If a committee of 3 people is to be selected from among 5 [#permalink]
Show Tags
02 Dec 2014, 08:37
vscid wrote: I understand the 1x approach, but if I were to do it the straighforward way, I get
10 x 8 x 6 (first place 10 ways, second place 8 ways, third place 6 ways) = 480, which is wrong.
What am I missing here? You missed the ordering. Since there are 6 different ways for these 3 people committee to be ordered . so divide 480/6= 80.




Re: If a committee of 3 people is to be selected from among 5
[#permalink]
02 Dec 2014, 08:37



Go to page
1 2
Next
[ 34 posts ]



