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If a committee of 3 people is to be selected from among 5
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05 Jan 2010, 07:46
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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: committee of 3
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05 Jan 2010, 10:07
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: committee of 3
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05 Jan 2010, 08:06
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: committee of 3
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05 Jan 2010, 09:03
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: committee of 3
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06 Jan 2010, 11:34
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: committee of 3
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06 Jan 2010, 11:37
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.



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Re: committee of 3
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07 Jan 2010, 05:30
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: committee of 3
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27 Feb 2010, 10:39
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?
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Re: committee of 3
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28 Sep 2010, 08:38
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.



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Re: committee of 3
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11 Oct 2010, 02:43
"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: committee of 3
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15 Jan 2011, 04:28
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 ??



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Re: committee of 3
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15 Jan 2011, 14:55
srivicool wrote: 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 ?? This issue is discussed here: pscombinations94068.html and here: iftherearefourdistinctpairsofbrothersandsisters99992.html
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Re: If a committee of 3 people is to be selected from among 5
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25 Oct 2013, 19:01
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))



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Re: If a committee of 3 people is to be selected from among 5
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26 Nov 2013, 00:53
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
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Re: committee of 3
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12 Dec 2013, 22:29
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)



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Re: If a committee of 3 people is to be selected from among 5
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24 Apr 2014, 13:48
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 ?



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Re: Combinatorics question
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22 May 2014, 22:26
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
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Re: If a committee of 3 people is to be selected from among 5
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15 Jul 2014, 04:06
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.
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If a committee of 3 people is to be selected from among 5
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15 Sep 2014, 18:02
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.



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Re: If a committee of 3 people is to be selected from among 5
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02 Dec 2014, 09:37
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
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02 Dec 2014, 09:37



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