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TheChakra
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Post subject: A committee of three people is to be chosen from four  Posted: Thu Oct 25, 2007 7:31 pm |
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A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve the committee?
16
24
26
30
32
This is how I calculated. Is this correct or I just got lucky?
I can basically select 3 people from 4 groups in 4C3 = 4 ways. I can select each of those 3 in 2 ways so total = 4x2x2x2 = 32 = official answer.
I tried solving using anagram method, but couldn't. Can someone confirm my method or show how to solve using anagram?
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raulverde
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Post subject: Posted: Sat Oct 27, 2007 3:51 pm |
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Lets say the couple are as follows :
Male 1 , Female 1
Male 2 , Female 2
Male 3 , Female 3
Male 4 , Female 4
For the 1st Member of the committee we have 8 choices (Assume we picked Male 1)
For the 2nd Member of the committee we have 6 choices (Already picked Male 1 and cant pick Female 1. Assume this time we picked Female 3 )
For the 3rd Member of the committee we have 4 choices (Already picked Male 1 and Female 3. Cant pick Female 1 and Male 3)
Now the committee can be arranged in 3! ways = 6
So the different committee are : (8*6*4)/6 = 32
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Nov1907
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Post subject: Posted: Mon Oct 29, 2007 1:32 pm |
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Total number of committees possible = 8C3 = 56.
Let there be 3 people A, B and C in the committee. Number of ways of selecting a committee of 3 with a married couple : A and B married while C is from the remaining 6 people. No: of ways of selecting people so that A and B are maried = 4. Number of Ways odf selecting C = 6C1 = 6. Therefore number of committees with married members = 4*6C1= 24.
So number of committees with no married members = 56 - 24 = 32
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RonPurewal
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Post subject: Posted: Tue Oct 30, 2007 4:11 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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Your original solution is completely valid, so long as you replace the phrase 'I can basically select 3 people from 4 groups in 4C3 = 4 ways' with 'I can basically select 3 of 4 groups in 4C3 = 4 ways.'
It's not a clean anagram, but you can do it with letters in the following way:
Let the couples be Aa, Bb, Cc, Dd. The committee can comprise anyone, so long as it doesn't contain the capital and lowercase versions of the same letter. So you have to have three different letters, but capital and lowercase are totally arbitrary.
* There are four ways to pick the 3 different letters (analogous to your 4C3): abc, abd, acd, bcd.
* In each group there are 8 ways to pick capital or lowercase (analogous to your 2x2x2): ABC ABc AbC aBC Abc aBc abC abc.
* 4 x 8 = 32.
This is essentially your solution, but in more intuitive terms.
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PermuationProb
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Post subject: Need to see anagram method Posted: Mon May 12, 2008 10:58 pm |
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Can someone please show me how to do this the MGMAT Anagram method? Sorry for being the dumb one.
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RonPurewal
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Post subject: Re: Need to see anagram method Posted: Wed May 14, 2008 5:43 am |
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Posts: 7146
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PermuationProb wrote: Can someone please show me how to do this the MGMAT Anagram method? Sorry for being the dumb one.
you can't do this problem by a pure anagram, because the pure anagram method doesn't allow for the exclusion of certain combinations. so if you did an anagram, you'd find the total number of combinations of any three of the people, including forbidden pairings.
you could do the anagram anyway - which would be YYYNNNNN, giving (8!)/(3!5!) - but then you'd have to subtract out all the possibilities involving forbidden pairings. and how would you find those, you ask? well... you'd basically have to list them. and that is certainly worse than the solutions listed on this page.
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Guest
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Post subject: A committee of three people is to be chosen from four Posted: Fri Aug 08, 2008 4:33 pm |
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Hi,
Can you please tell me what is wrong with my approach ?
First I calculated the total number of ways of selecting 3 people from 8 i.e 8C3 = 56
Then I took the reverse gear and tries to calculate the committee of 3 people by filing up the three places __ __ __ such that the first place can be taken by any of the 8 members , second has to be taken by the first person's partner so there is only one possibilty and the thrid place can be taken by any of the 6 members... So the total number of ways = 8*1*6 = 48
So the number of different committees with the same partner = 56-48 = 8 , So does it mean that I have to repeat this process 4 times as in multiply this answer by 4 to get 32 because there are 4 such committees... I usually get confused about when to multiply and when not to...Thnx
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RonPurewal
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Post subject: Re: A committee of three people is to be chosen from four Posted: Thu Aug 14, 2008 4:23 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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the problem is here:
Anonymous wrote: Then I took the reverse gear and tries to calculate the committee of 3 people by filing up the three places __ __ __ such that the first place can be taken by any of the 8 members , second has to be taken by the first person's partner so there is only one possibilty and the thrid place can be taken by any of the 6 members... So the total number of ways = 8*1*6 = 48
this is good, but then you have to divide this result by 2, because you've counted every one of these committees twice.
here's why: when you say 8 for the first possibility, that includes both people in each couple. therefore, considering one particular couple - let's call them couple "A" - you're going to count each committee once by starting with husband "A", and then again by starting with wife "A". therefore, you should divide by 2.
so, 48/2 = 24 --> 56 - 24 = 32.
--
incidentally, if you're going to go this route, there is a much easier way to calculate the total # of committees involving one of the married couples: just take the number of couples (4) times the number of possible 3rd persons (6). that's 4 x 6 = 24, so 56 - 24 = 32.
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himanshubari
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Post subject: Re: A committee of three people is to be chosen from four Posted: Wed Aug 26, 2009 7:56 pm |
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Posts: 4
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I was trying to solve this problem using the counting method.
there are 4 couples so total 8 people
We need a 3 member committee where none of the 3 are a couple
SO 1st person can be chosen in 8 ways
2nd person can be chosen in 8-2 = 6 ways
and the 3rd person can be chosen in 8-2-2 = 4 ways
But my method gets me 8*6*4 which is wrong..
What am I missing here?
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himanshubari
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Post subject: Re: A committee of three people is to be chosen from four Posted: Wed Aug 26, 2009 8:01 pm |
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Posts: 4
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himanshubari wrote: I was trying to solve this problem using the counting method.
there are 4 couples so total 8 people
We need a 3 member committee where none of the 3 are a couple
SO 1st person can be chosen in 8 ways
2nd person can be chosen in 8-2 = 6 ways
and the 3rd person can be chosen in 8-2-2 = 4 ways
But my method gets me 8*6*4 which is wrong..
What am I missing here? OOPS.. got it had to divide by the permutatins possible as this is a combinations problem
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prashant.jakhetiya
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Post subject: Re: A committee of three people is to be chosen from four Posted: Fri Sep 18, 2009 6:25 am |
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Male --> A B C D
Female--> 1 2 3 4
The committee of 3 out of 4
Case 1>
a>Take only males 4C3
b>Take only females 4C3
Case 2>
a> Two from Four Males 4C2 (Suppp A and B )& One from the remaining 2 Females (3 and 4) 2C1 = 4C2 * 2C1
b> Two from Four Females 4C2 & One from the remaining 2 Males 2C1
Add Case1 and Case2
4C3+4C3+4C2 *2C1 + 4C2 *2C1 = 32 .
Does this make sense .
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RonPurewal
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Post subject: Re: A committee of three people is to be chosen from four Posted: Sat Sep 26, 2009 2:24 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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himanshubari wrote: I was trying to solve this problem using the counting method.
there are 4 couples so total 8 people
We need a 3 member committee where none of the 3 are a couple
SO 1st person can be chosen in 8 ways
2nd person can be chosen in 8-2 = 6 ways
and the 3rd person can be chosen in 8-2-2 = 4 ways
But my method gets me 8*6*4 which is wrong..
What am I missing here? yes, this is the simplest way. you should be able to solve any combinatorics problem with the "slot" (multiplication) method.you may have to divide by a factorial or factorials (if order doesn't matter), but, other than that, the slot method should pretty much always work. in this problem: as the poster has noted above, there are 8 choices for the first person, 6 for the second, and 4 for the third. but ORDER DOESN'T MATTER, so you have to divide by the FACTORIAL OF THE NUMBER OF THINGS CHOSEN: (8 x 6 x 4) / 3! = 32
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RonPurewal
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Post subject: Re: A committee of three people is to be chosen from four Posted: Sat Sep 26, 2009 2:26 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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prashant.jakhetiya wrote: Male --> A B C D
Female--> 1 2 3 4
The committee of 3 out of 4
Case 1>
a>Take only males 4C3
b>Take only females 4C3
Case 2>
a> Two from Four Males 4C2 (Suppp A and B )& One from the remaining 2 Females (3 and 4) 2C1 = 4C2 * 2C1
b> Two from Four Females 4C2 & One from the remaining 2 Males 2C1
Add Case1 and Case2
4C3+4C3+4C2 *2C1 + 4C2 *2C1 = 32 .
Does this make sense . this does work. although there is no way that anybody who struggles with combinatorics is going to come up with it in two minutes. see the previous post for a MUCH shorter solution to this problem.
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sudaif
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Post subject: Re: A committee of three people is to be chosen from four Posted: Tue Aug 03, 2010 12:30 pm |
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Posts: 126
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RonPurewal wrote: himanshubari wrote: I was trying to solve this problem using the counting method.
there are 4 couples so total 8 people
We need a 3 member committee where none of the 3 are a couple
SO 1st person can be chosen in 8 ways
2nd person can be chosen in 8-2 = 6 ways
and the 3rd person can be chosen in 8-2-2 = 4 ways
But my method gets me 8*6*4 which is wrong..
What am I missing here? yes, this is the simplest way. you should be able to solve any combinatorics problem with the "slot" (multiplication) method.you may have to divide by a factorial or factorials (if order doesn't matter), but, other than that, the slot method should pretty much always work. in this problem: as the poster has noted above, there are 8 choices for the first person, 6 for the second, and 4 for the third. but ORDER DOESN'T MATTER, so you have to divide by the FACTORIAL OF THE NUMBER OF THINGS CHOSEN: (8 x 6 x 4) / 3! = 32 thanks Ron!
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RonPurewal
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Post subject: Re: A committee of three people is to be chosen from four Posted: Thu Aug 05, 2010 9:34 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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sudaif wrote: RonPurewal wrote: himanshubari wrote: I was trying to solve this problem using the counting method.
there are 4 couples so total 8 people
We need a 3 member committee where none of the 3 are a couple
SO 1st person can be chosen in 8 ways
2nd person can be chosen in 8-2 = 6 ways
and the 3rd person can be chosen in 8-2-2 = 4 ways
But my method gets me 8*6*4 which is wrong..
What am I missing here? yes, this is the simplest way. you should be able to solve any combinatorics problem with the "slot" (multiplication) method.you may have to divide by a factorial or factorials (if order doesn't matter), but, other than that, the slot method should pretty much always work. in this problem: as the poster has noted above, there are 8 choices for the first person, 6 for the second, and 4 for the third. but ORDER DOESN'T MATTER, so you have to divide by the FACTORIAL OF THE NUMBER OF THINGS CHOSEN: (8 x 6 x 4) / 3! = 32 thanks Ron! glad it helped
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