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If a committee of 3 people is to be selected from among 5 married coup
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05 Jan 2010, 06:46
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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
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Re: If a committee of 3 people is to be selected from among 5 married coup
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05 Jan 2010, 09:07
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.
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Re: If a committee of 3 people is to be selected from among 5 married coup
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05 Jan 2010, 07:06
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)
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Re: If a committee of 3 people is to be selected from among 5 married coup
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05 Jan 2010, 08:03
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




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Re: If a committee of 3 people is to be selected from among 5 married coup
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06 Jan 2010, 10:34
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.



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Re: If a committee of 3 people is to be selected from among 5 married coup
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07 Jan 2010, 04:30
I too got 80 with the conventional way of 10C3  5C1 * 8C1 = 120  40 = 80. But learnt and loved Bunuel's way. Thanks!



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Re: If a committee of 3 people is to be selected from among 5 married coup
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04 Aug 2010, 03:20



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Re: If a committee of 3 people is to be selected from among 5 married coup
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04 Aug 2010, 03:26
Looking at the couples at first as single units was the eyeopener. Thanks very kindly.



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Re: If a committee of 3 people is to be selected from among 5 married coup
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11 Oct 2010, 01:43
"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



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Re: If a committee of 3 people is to be selected from among 5 married coup
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29 Feb 2012, 10:01
5C3 > ways 3 couples can be selected out of 5 couples = 10 (2C1)^3 > ways to select 1 adult from each couple, for each of the three couples. = 8 Total Possibilities = 5C3 * (2C1)^3 = 80



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Re: If a committee of 3 people is to be selected from among 5 married coup
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29 Feb 2012, 10:16
I'm pretty lost on this one..
Here was the logic that i thought would work but obviously I am missing something:
5 couples = 10 people
So first position there are 10 options
Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options
3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options
so I got 10 * 9 * 8 = 480
What am I missing?



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Re: If a committee of 3 people is to be selected from among 5 married coup
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Updated on: 29 Feb 2012, 10:26
AbeinOhio wrote: I'm pretty lost on this one..
Here was the logic that i thought would work but obviously I am missing something:
5 couples = 10 people
So first position there are 10 options
Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options
3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options
so I got 10 * 9 * 8 = 480
What am I missing? You have to select a committee of 3 people. It means A, B, C form the same committee as B, C, A i.e. you don't have to arrange people. But when you say 'first person' can be selected in 10 ways, you are arranging the 3 people in first, second and third spots. So according to you, A, B, C and B, C, A are different committees. So all you need to do in unarrange. To arrange 3 people, you multiply by 3! To unarrange, you will divide by 3! 480/3! = 80 There's your answer. Check out this post for a detailed explanation: http://www.veritasprep.com/blog/2011/11 ... nstraints/
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Re: If a committee of 3 people is to be selected from among 5 married coup
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29 Feb 2012, 10:25
AbeinOhio wrote: I'm pretty lost on this one..
Here was the logic that i thought would work but obviously I am missing something:
5 couples = 10 people
So first position there are 10 options
Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options
3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options
so I got 10 * 9 * 8 = 480
What am I missing? You've done everything right, just missed the last step: since the people in the group needn't be ordered (arranged) in it, then you should divide 10*8*6=480 by 3! to get rid of duplication and get unordered groups > 480/3!=80. Hope it's clear.
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Re: If a committee of 3 people is to be selected from among 5 married coup
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05 Nov 2012, 21:51
Bunuel wrote: kwhitejr wrote: Can anyone demonstrate the following?
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 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 chosen couples can send two persons (either husband or wife): 2*2*2=2^3=8. Total # of ways: 5C3*2^3=80. Answer: D. Similar problems: pscombinations94068.htmlpscombinations101784.htmlcommitteeof88772.htmlif4peopleareselectedfromagroupof6marriedcouples99055.htmliftherearefourdistinctpairsofbrothersandsisters99992.htmlHope it's clear. How it could be two person, out of three couples they have to sent 3 rep rite.. I am totally confused kindly enlighten me



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Re: If a committee of 3 people is to be selected from among 5 married coup
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06 Nov 2012, 02:43
breakit wrote: Bunuel wrote: kwhitejr wrote: Can anyone demonstrate the following?
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 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 chosen couples can send two persons (either husband or wife): 2*2*2=2^3=8. Total # of ways: 5C3*2^3=80. Answer: D. Similar problems: pscombinations94068.htmlpscombinations101784.htmlcommitteeof88772.htmlif4peopleareselectedfromagroupof6marriedcouples99055.htmliftherearefourdistinctpairsofbrothersandsisters99992.htmlHope it's clear. How it could be two person, out of three couples they have to sent 3 rep rite.. I am totally confused kindly enlighten me Yes, 3 couples should send total of 3 representatives, but EACH couple has 2 options (either husband or wife), thus these 3 couples can send 3 persons in 2^3=8 ways. Hope it's clear.
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Re: If a committee of 3 people is to be selected from among 5 married coup
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27 Dec 2012, 18:27
kwhitejr 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 I always use the ANAGRAM technique. How many ways can we select three persons from 5 couples? Since we can only get one representative from each couple selected, we can imagine the 5 couples as 5 persons. 5!/3!2! = 10 Now, we know that there are 10 ways to select three persons of representing couples. But whenever we make selections, there are two ways to select from the couple, either wife or husband. Thus, 2 x 2 x 2 = 8 8*10 = 80 Answer: C



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Re: If a committee of 3 people is to be selected from among 5 married coup
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07 Jan 2013, 10:43
5C3  select three couples 2*2*2 > select one member from each couple
ans  5C3 * 8 = 80



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Re: If a committee of 3 people is to be selected from among 5 married coup
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07 Jan 2013, 10:52
Find the number of ways you can have a couple in the committee and then subtract from the total number of 3person committees. 10C3 = 120 possible combinations Since there are 5 couples and the final spot could be filled with any of the 8 remaining people you have 8x5 = 40 ways to achieve this. Therefore you have 12040 = 80 committees w/out a married couple. Answer: D



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Re: If a committee of 3 people is to be selected from among 5 married coup
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15 Jul 2013, 08:11
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?
(1) Combinatorial approach:
\(C^5_3*(C^2_1)^3 =\frac{5*4*3}{3*2}* 2^3 = 10*8 = 80\)
(2) Reversal combinatorial approach:
Total number of groups: \(C^10_3 = 120\) Total number of groups with married people: \(C^5_1 * C^4_1 * C^2_1 = 5*4*2 = 40\)
120  40 = 80



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Re: If a committee of 3 people is to be selected from among 5 married coup
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15 Jul 2013, 17:44
Maxirosario2012 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?
(1) Combinatorial approach:
\(C^5_3*(C^2_1)^3 =\frac{5*4*3}{3*2}* 2^3 = 10*8 = 80\) i.e., IF AB, CD, EF, GH, IJ are the couples you have to select only one from a group 1. So selecting one from a group of 2 can be done in \(2C1\)ways. i.e., in 2 ways You have to select 3 people that way. So the total number of possibilities is \(2*2*2 = 8\) 2. Each group of 2 itself has to be selected from 5 such groups. You are selecting 3 groups of 2 from 5 such groups. therefore the total number of possibilities for this is \(5C3= 10\). 3. \((1) * (2) = 8* 10 = 80.\) Quote: (2) Reversal combinatorial approach:
Total number of groups: \(C^10_3 = 120\) Total number of groups with married people: \(C^5_1 * C^4_1 * C^2_1 = 5*4*2 = 40\)
120  40 = 80 1. Total number of possibilities of selecting 3 people out of 10 people\(= 10C3= 120\) 2. For total number of groups with a married couple the situation is you have (i)2 people who are married to each other i.e., a group of 2 married to each other and (ii) one other person. The number of possibilities of (i) is the number of ways one group of married people can be selected from 5 groups of married people which is \(5C1 = 5\)ways, the number of possibilities of (ii) can be arrived by finding out in how many ways the remaining person can be selected. It can be done in \(8C1=8\) ways because if you remove the selected married couple 8 persons will remain. 3. Total number of ways of having a married couple\(in the group of 3 = 5*8=40\)4. So number of groups in which 2 people are not married couple \(= 12040=80.\)
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