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Re: A committee of 3 people is to be chosen from four married [#permalink]
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07 Nov 2012, 05:37



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Re: A committee of 3 people is to be chosen from four married [#permalink]
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27 Dec 2012, 21:54
LM wrote: A committee of 3 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 on the committee?
A. 16 B. 24 C. 26 D. 30 E. 32
How many ways to select 3 represented couples from 4 couples? 4!/3!1! = 4 How many ways to select a person from a pair? 2 \(=4 * 2 * 2 * 2 = 32\) Answer: E More detailed explanation here : Selection/Deselection Technique
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Re: A committee of 3 people is to be chosen from four married [#permalink]
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02 Sep 2013, 23:55
I solved it using this method, hope I'm using the correct concept
total 4 couples = 8 people in total total no of ways to choose 3 people out of 8 = 8!/(5!3!) = 56 No. of ways couples are included in the com = 4! = 24 Therefore no. of couples with no couples included = 5624 = 32.



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Re: Problem Solving Question [#permalink]
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26 Oct 2013, 19:03
schokshi99 wrote: 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 on the committee?
A. 16 B. 24 C. 26 D. 30 E. 32
Please explain the answer. Because people who are married to each other cannot both serve on the committee, three members of the committee must come from 3 different couples. We have 4 ways to pick 3 couples from 4 couples. For each couple, we have 2 choices to pick one of them. So the answer is 4 * 2 * 2 *2 = 32 (E)
Last edited by tuanle on 26 Oct 2013, 21:39, edited 2 times in total.



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Re: Problem Solving Question [#permalink]
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26 Oct 2013, 21:20
Agreed with Tuanle.
Another easy way is to write down all the ways to select
We have total 4 men and 4 women from which we need to form a committee of 3 such that husband and wife wont be included
The combinations are
1) MMM (all 3 men)  4c3 = 4 2) MMW (Two men and one women)  4c2 * 2c1 (the wives of selected men should not be included) = 12 3) WWM  4c2*2c1 = 12 4) WWW  4c3 = 4
Total = 4+12+12+4=32



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Re: Problem Solving Question [#permalink]
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27 Oct 2013, 05:47



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Re: A committee of 3 people is to be chosen from four married [#permalink]
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12 Nov 2013, 11:42
instead of thinking as husband and wife in the group, think of them as all different ppl in groups of two. you can only select one person from each group
so that means
(a,b) (c,d) (e,f) (g,h)
and we need to fill in 3 spaces
_ x_ x_
any one person can come from each group, so three spaces filled by one person from each group are
2x2x2  (A)
taking different combos of the group => 4C3 => 4  (B)
multiply (A) and (B) gives all the different ways this group can be created
(A) x (B) = 2x2x2x4 = 32 Thus choice E



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Re: PSCombinations [#permalink]
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23 Apr 2014, 20:01
Bunuel wrote: A committee of 3 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 on the committee? A. 16 B. 24 C. 26 D. 30 E. 32
One of the approaches:
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) out of 4 to send only one "representatives" to the committee: 4C3=4.
But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.
Total # of ways: 4C3*2^3=32.
Answer: E. Hi Bunuel, Sorry for the tedious questions. I know this has been addressed in the other posts but still having a hard time grasping the concept, if I use the concept of 8(first choice) x6(second choice)x 4(third choice) ...why do I need to divide by 3!. Doesn't that mean that there are orders in how these members are getting chosen and we don't care for that? meaning, it's a permutation problem  am i correct? IF it is a permutation problem, why aren't we using the traditional permutation formula of 8!/3!? I also used another approach of (8c1)(6c1)(4c1) / (8c3). Why is that wrong? Conversely, if I use your method listed above, I can get onboard with the "4c3" part as we need 3 different couples out of the 4 but i'm not grasping the concept of cubing "2". Shouldn't we be doing 2^4? Thanks a ton.



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Re: A committee of 3 people is to be chosen from four married [#permalink]
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23 Apr 2014, 20:31
Bunuel wrote: watwazdaquestion wrote: is this a correct way to get the answer? or was it just coincidence:
8C3  4(4C2) = 56  4(6) = 32 It's not clear what is the logic behind the formula. Reversed approach would be: There are 8C3=56 ways to select 3 people out of 8 without any restriction; There are 4C1*6=24 ways there to be a couple among 3 members: 4C1 ways to select a couple out of 4, which will be in the committee and 6 ways to select the third remaining member (since there will be 6 members left after we select a couple out of 8 people). 5624=32. Hope it's clear. Hi Bunuel, I'm confused by this step, which is also outlined above. "There are 4C1*6=24 ways there to be a couple among 3 members:". Can you please elaborate on this?



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Re: A committee of 3 people is to be chosen from four married [#permalink]
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24 Apr 2014, 01:25
russ9 wrote: Bunuel wrote: watwazdaquestion wrote: is this a correct way to get the answer? or was it just coincidence:
8C3  4(4C2) = 56  4(6) = 32 It's not clear what is the logic behind the formula. Reversed approach would be: There are 8C3=56 ways to select 3 people out of 8 without any restriction; There are 4C1*6=24 ways there to be a couple among 3 members: 4C1 ways to select a couple out of 4, which will be in the committee and 6 ways to select the third remaining member (since there will be 6 members left after we select a couple out of 8 people). 5624=32. Hope it's clear. Hi Bunuel, I'm confused by this step, which is also outlined above. "There are 4C1*6=24 ways there to be a couple among 3 members:". Can you please elaborate on this? First couple: \(A_1,A_2\); Second couple: \(B_1,B_2\); Third couple: \(C_1,C_2\); Fourth couple: \(D_1,D_2\). We want to select 3 people: a couple and one more. We can select any from 4 couples (4 options) and for the third member we can select any from the remaining 6 people. For example if we select \(A_1,A_2\), then we can select third member from \(B_1,B_2,C_1,C_2,D_1,D_2\). Hope it's clear.
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Re: A committee of 3 people is to be chosen from four married [#permalink]
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24 Apr 2014, 19:12
Bunuel wrote:
First couple: \(A_1,A_2\); Second couple: \(B_1,B_2\); Third couple: \(C_1,C_2\); Fourth couple: \(D_1,D_2\).
We want to select 3 people: a couple and one more.
We can select any from 4 couples (4 options) and for the third member we can select any from the remaining 6 people. For example if we select \(A_1,A_2\), then we can select third member from \(B_1,B_2,C_1,C_2,D_1,D_2\).
Hope it's clear.
Hi Bunuel, I should've elaborated: If the goal is to find combinations with NO couples, how does "totalcombo of at least 1 couple" equal "no couples"? Aren't there possibilities of 2,3,4 couples? Additionally, why wouldn't it be (4c1)(2^3)? Is this equation saying that we should chose 1 couple out of 4 and we have 3 couples left so 2*2*2?



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Re: A committee of 3 people is to be chosen from four married [#permalink]
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06 May 2014, 09:06
russ9 wrote: Bunuel wrote:
First couple: \(A_1,A_2\); Second couple: \(B_1,B_2\); Third couple: \(C_1,C_2\); Fourth couple: \(D_1,D_2\).
We want to select 3 people: a couple and one more.
We can select any from 4 couples (4 options) and for the third member we can select any from the remaining 6 people. For example if we select \(A_1,A_2\), then we can select third member from \(B_1,B_2,C_1,C_2,D_1,D_2\).
Hope it's clear.
Hi Bunuel, I should've elaborated: If the goal is to find combinations with NO couples, how does "totalcombo of at least 1 couple" equal "no couples"? Aren't there possibilities of 2,3,4 couples? How there be more than one couple in 3 people?
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Re: A committee of 3 people is to be chosen from four married [#permalink]
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18 May 2014, 22:22
No specific information on the Guys Vs Gals in the committee 3 Guys: 4C3 2 Guys 1 Girl: 4C2 * 2C1> Choose from the girls who are not married to the 2 guys chosen already 3Girls: 4C3 2 Girls 1 Guy: 4C2 * 2C1> Choose from the guys who are not married to the 2 girls chosen already 4+4+12+12=32
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Re: A committee of 3 people is to be chosen from four married [#permalink]
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02 Jul 2014, 00:49
First, let us select the couple 1, so if H1W1 is selected there can there can be only 6 ways of choosing 1 member from the other 3 couples. It goes the same for the other three couples. So, if we are to select the members such that there would always be a couple in the committee then it can be done in 6 *4 ways= 24 ways.
Now we can select any 3 members from 8 persons in 8C3 =56 ways.
So no of ways of selecting members who are not couples is 5624=32 ways



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Re: A committee of 3 people is to be chosen from four married [#permalink]
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12 Nov 2014, 23:44
8*6*4. then divide by 3! to get rid of duplicates



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Re: A committee of 3 people is to be chosen from four married [#permalink]
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23 Feb 2016, 05:25
can we solve like this way
8C3 total 56 committees and now we gave to select only male combination is 4! 56  4! = 32



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A committee of 3 people is to be chosen from four married [#permalink]
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23 Feb 2016, 05:33
pkk1611 wrote: can we solve like this way
8C3 total 56 committees and now we gave to select only male combination is 4! 56  4! = 32 hi, the answer may be correct but the method/logic is not clear.. why are you subtracting male combinations?
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Re: A committee of 3 people is to be chosen from four married [#permalink]
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02 Apr 2016, 12:11
shrouded1 wrote: So there are 3 people we need to choose from 8.
Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways
Total ways = 4+4+12+12 = 32
Posted from my mobile device wonderful explanation . I really liked .



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Re: A committee of 3 people is to be chosen from four married [#permalink]
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25 Jun 2016, 16:37
We can solve this in two general steps.
1) We can pretend order matters (which can be much easier to visualize) 2) Then we can think about what happens if order doesn't matter.
1) If order matters, we can visualize three chairs or slots or whatever and consider them one at a time: Anybody can sit in the first chair, so there are 8 options for chair one. But now that somebody is in chair one, there are only 6 options for chair two (it can't be the person in chair one and it can't be that person's spouce). And for every set of people occupying chairs one and two, there are only 4 remaining people allowed to sit in chair three.
That means that there are 8*6*4 ways to fill the three chairs.
2) But we're looking for 'committees.' and for committees order doesn't matter (whereas it does when we're visualizing chairs). For example, Andy, Beth, Charlie would be the same committee as Beth, Andy Charlie.
In fact there are six ways that Andy Beth and Charlie could fill the chairs:
ABC ACB BAC BCA CAB CBA
So for each committee, we counted 6 different ways for them to fill the chairs.
In other words, our 8*6*4 number is too big by a factor of 6.
Therefore, the answer is (8*6*4)/6 = 32




Re: A committee of 3 people is to be chosen from four married
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