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If a committee of 3 people is to be selected from among 5 [#permalink]
 05 Jan 2010, 07:46
Question Stats:
81% (01:39) correct
18% (01:03) wrong based on 2 sessions
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 [#permalink]
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 120-40=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 [#permalink]
05 Jan 2010, 08:37
i still didn't get the second part! 
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Re: committee of 3 [#permalink]
05 Jan 2010, 09:03
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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=120-40=80 hence D
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Re: committee of 3 [#permalink]
05 Jan 2010, 09:16
Ahh got it, thanks to both of u 
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Re: committee of 3 [#permalink]
05 Jan 2010, 10:07
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Re: committee of 3 [#permalink]
06 Jan 2010, 11:34
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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 [#permalink]
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 [#permalink]
07 Jan 2010, 05:30
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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 [#permalink]
27 Feb 2010, 10:39
I understand the 1-x 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 [#permalink]
28 Sep 2010, 04:31
Nice post ..Thanks
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Re: committee of 3 [#permalink]
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 [#permalink]
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 [#permalink]
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 [#permalink]
15 Jan 2011, 14:55
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Re: If a committee of 3 people is to be selected from among 5 [#permalink]
11 Jun 2012, 22:20
this is a great post thanks
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Re: If a committee of 3 people is to be selected from among 5
[#permalink]
11 Jun 2012, 22:20
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