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If a committee of 3 people is to be selected from among 5

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If a committee of 3 people is to be selected from among 5 [#permalink] New post 05 Jan 2010, 06:46
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A
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D
E

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67% (01:47) correct 32% (01:13) wrong based on 168 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
[Reveal] Spoiler: OA
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Re: committee of 3 [#permalink] New post 05 Jan 2010, 07:06
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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] New post 05 Jan 2010, 08:03
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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] New post 05 Jan 2010, 09:07
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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?

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.
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Re: committee of 3 [#permalink] New post 06 Jan 2010, 10:34
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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] New post 06 Jan 2010, 10:37
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] New post 07 Jan 2010, 04:30
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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] New post 27 Feb 2010, 09:39
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] New post 28 Sep 2010, 07:38
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] New post 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: committee of 3 [#permalink] New post 15 Jan 2011, 03:28
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] New post 15 Jan 2011, 13:55
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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: ps-combinations-94068.html and here: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 11 Jun 2012, 21:20
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] New post 21 Aug 2013, 16:36
Can someone please help me? I don't know what I am doing wrong.

5C1*2C1*4C1*2C1*8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple
4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple
8C1 - people that can occupy the third spot
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 22 Aug 2013, 02:23
Expert's post
brunawang wrote:
Can someone please help me? I don't know what I am doing wrong.

5C1*2C1*4C1*2C1*8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple
4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple
8C1 - people that can occupy the third spot


The following post might help: a-five-member-committee-is-to-be-formed-from-a-group-of-five-55410.html#p1248236
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Re: committee of 3 [#permalink] New post 26 Aug 2013, 08:38
mrblack wrote:
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.


i don't understand. We need one more member, so we choose 1 “couple"? (hence 2 members)
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Re: committee of 3 [#permalink] New post 27 Aug 2013, 03:12
Expert's post
oss198 wrote:
mrblack wrote:
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.


i don't understand. We need one more member, so we choose 1 “couple"? (hence 2 members)


It's in the same line as: if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html#p669797
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 25 Oct 2013, 18:01
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 [#permalink] New post 25 Nov 2013, 23:53
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 [#permalink] New post 12 Dec 2013, 21:29
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 (un-arranged the order as it doesn't matter)
Re: committee of 3   [#permalink] 12 Dec 2013, 21:29
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