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# PS-Combinations

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PS - Combination [#permalink]  18 Nov 2007, 20:15
A committee of three people is to be chosen from 4 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?

16
24
26
30
32

Here's my solution so far :

Total combination 8C3 = 56

Choose the first couple 4C1 = 4
Choose the third person= 8C1 = 8 * 7! / 7! = 8

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I would say the answer is 32:

1st member: one of the 8, so we have 8 candidates for the 1st member
2nd: we have only 6 cand.
3rd - we have 4 cand.

So totally we have (8 * 6 * 4) / 3! (we devide by 3! as order does not metter)
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Following the original poster's approach, I think the # of ways to choose the third person is 6C1.

As the first and second people is selected from the married couples, that only leaves 6 people ... out of which any 1 can be chosen
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E. 32

Total no of ways of choosing a committee = 8C3 = 56

No of ways to choose first couple = 4C1 = 4
No of ways to choose 1 person from remaining 6 = 6C1 = 6

Total number of ways of choosing a married couple to be in the same committee = 4*6=24

So, no of ways not choosing a married couple = 56-24 =32
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Re: PS - Combination [#permalink]  20 Nov 2007, 13:50
A committee of three people is to be chosen from 4 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?

16
24
26
30
32

Here's my solution so far :

Total combination 8C3 = 56

Choose the first couple 4C1 = 4
Choose the third person= 8C1 = 8 * 7! / 7! = 8

Total combination 8C3 = 56
Choose the first couple 4C1 = 4
Choose the third person= 6C1 = 6
so # of comittees with a couple = 4x6 = 24
so # of comittees without a couple = 56 - 24 = 32
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Another solution (you can check yourself in this way):
Choose 3 couples out of 4 C(4,3)=4
Then choose one person from each couple: 4*2*2*2=32 -E it is
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Re: PS - Combination [#permalink]  27 Sep 2009, 23:35
A committee of three people is to be chosen from 4 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?

16
24
26
30
32

Soln:
Now the first of the 3 people commitee can be chosen from any of the 8 people in 8 ways
Now the second of the 3 people commitee can be chosen from any of the 6 people in 6 ways(leaving out the one person who is the couple of the person chosen in the first statement)
Now the third of the 3 people commitee can be chosen from any of the 4 people in 4 ways(leaving out the two people whose couples have already been chosen)

Thus the total number of permutations is = 8 * 6 * 4
But since this is a selection, hence order does not matter

Therefore the total number of selections is = 8 * 6 * 4/3!
= 32 ways
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Re: PS - Combination [#permalink]  28 Dec 2009, 20:13
yep! same result .... 4C3*(2C1)^3 = 32

4C3: Select 3 couples out of 4
2C1: from each couple, select 1 person not married
(2c1)^3: there are 3 couples selected
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PS-Combinations [#permalink]  11 May 2010, 12:35
Sorry...I am not able to move this post to PS section. Could someone please move it to PS section.

Approach I thought is as follows...if some shorter method is possible please explain..

total selections = 8C3 = 56

let's say that couple is always present in this committee of three.

This means that there are 4 ways to select 2 people of the committee. ( 4 couples and any one couple can be selected in 4 ways)
The third person can be selected out of remaining 6 people in 6 ways.

Therefore when couple exists there are: 4X6 = 24 ways

Thus no couple = 8C3 - (4X6) = 32
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Re: PS-Combinations [#permalink]  11 May 2010, 15:24
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.

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Re: PS-Combinations [#permalink]  17 Jun 2010, 02:16
Hi everybody,
I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.

i can choose first member of the comitee in 8 ways..
removing the spouse of the selected person second member can be chosen in 6 ways....
third member in 4 ways.....
so 8*6*4 which is not answer can someone explain why?
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Re: PS-Combinations [#permalink]  17 Jun 2010, 05:50
1
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amitjash wrote:
Hi everybody,
I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.

i can choose first member of the comitee in 8 ways..
removing the spouse of the selected person second member can be chosen in 6 ways....
third member in 4 ways.....
so 8*6*4 which is not answer can someone explain why?

The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people - 3! --> 192/3!=32.

Consider this: there are two couples and we want to choose 2 people not married to each other.
Couples: A_1, A_2 and B_1, B_2. Committees possible:

A_1,B_1;
A_1,B_2;
A_2,B_1;
A_2,B_2.

Only 4 such committees are possible.

If we do the way you are doing we'll get: 4*2=8. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

You can see the similar problem at:
committee-of-88772.html#p669797

Hope it helps.
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Q from gmatprep2 [#permalink]  22 Jul 2010, 06:42
a committe of 3 ppl is to be chosen from 4 married couples. what are the no of different comittees that can be formed if 2 married ppl cant be on the committee

16
24
26
30
32
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Re: PS-Combinations [#permalink]  22 Jul 2010, 08:37
Bunuel wrote:

The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people - 3! --> 192/3!=32.

Thanks the two of you!

This is also the way I like to solve such questions and I believe it is way faster than any 10C3... and so on!
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Re: PS - Combination [#permalink]  27 Sep 2010, 06:52
1. total number of outcomes: 3C8=8!/3!5!=56
2. probability of the given event: 1*6/7*4/6=4/7
3. 56*4/7=32
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Re: PS - Combination [#permalink]  27 Sep 2010, 07:32
+1 for 32

Since you can't use 2 people from a single couple, you need to choose 3 different couples out of the 4. There are 4 ways to do this.

Then you choose 1 person from each of the 3 couples. There are (2*2*2) = 8 ways to choose the 3 people.

8 * 4 = 32
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Re: PS - Combination [#permalink]  27 Sep 2010, 07:53
Merging similar topics.

Also check similar question at: confuseddd-99055.html?hilit=married%20couples#p781169
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Re: PS - Combination [#permalink]  11 Oct 2010, 03:56
A committee of three people

Combo box arrangement
(_)(_)(_)/3!

is to be chosen from 4 married couples.

Bag of 8 to choose from.

What is the number of different committees that can be chosen

Start filling in combo box arrangement.
First position has 8 choices.
(8)(_)(_)/3!

if two people who are married to each other cannot both serve on the committee?

but the spouse gets eliminated, too.
Second position has 6 choices.
(8)(6)(_)/3!

but another spouse gets eliminated.
Third position has 4 choices.
(8)(6)(4)/3! = 8*4 = 32
Re: PS - Combination   [#permalink] 11 Oct 2010, 03:56
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