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In how many ways to choose a group of 3 people from 6 couples so that

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In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 10 Aug 2006, 18:14
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In how many ways to choose a group of 3 people from 6 couples so that no couple is chosen
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 10 Aug 2006, 18:28
Let me try....

6 couples = 12 people.

There are three slots to be filled.

If we start with any of the 12, the second person will have to be out of 10 because the spouse of the first one chosen must be left out. Same philosophy for the third person brings it to 8.

Therefore, 12*10*8 = 960
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 10 Aug 2006, 19:30
mand-y wrote:
in how many ways to choose a group of 3 people from 6 couples so that no couple is chosen


totoal = 12c3 = 220
2 couples and a single = 6x10 = 60
so non couple = 220-60 = 160
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 10 Aug 2006, 21:35
Hey professor,

I know you can probably help me with this one. Here is my approach toward the question:

Total possibilites:
12C3= 220

Couples:
6C3= 20

220-20 = 200

Could you explain why this is wrong?
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 10 Aug 2006, 22:23
160

Number of ways 3 people can be chosen from 12 = 12C3 = 220

Number of ways a couple can be chosen = 6
Number of ways the third person can be chosen = 10C1 = 10

Hence total number of ways a couple with another person can be chosen = 6*10 = 60

Therefore total number of ways a couple cannot be chosen = 220-60 = 160
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 11 Aug 2006, 03:21
Now i'm confused, I can see you need the formula 12c3 to start with, but can somebody explain the fallacy in my logic above?

On some probability/combinations questions I use the slot method approach and some I use the formula straight, when should I use these methods?
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 11 Aug 2006, 06:08
Just to add to confusion, here is another way:

3 people out of 6 couples with no couple selectd

1. Select either 3 men or 3 women: 2*6C3=40
2. Select 1 man+2 women: 6*5C2=60
3. Select 1 woman+2men: 6*5C2=60
-----------------
Total: 40+60+60=160 ways
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 11 Aug 2006, 06:17
OK, I think the fallacy in my logic above is that I assumed it was a perrmutation rather then a combination. In my approach, XYZ is different from ZXY, even though you have the same set of 3 with no couples.

Since there were 6 couples to mitigate the permuations, (12*10*8)/6 = 160.
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 11 Aug 2006, 10:27
rdw28 wrote:
Hey professor,

I know you can probably help me with this one. Here is my approach toward the question:

Total possibilites:
12C3= 220

Couples: 6C3= 20

220-20 = 200

Could you explain why this is wrong?


here in red part above, you only selected 2 people (actually a couple) not 3 people. you need to select 3 people.

agsfaltex wrote:
I assumed it was a perrmutation rather then a combination


yes, you are correct.
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 11 Aug 2006, 12:23
Wow, I cannot believe I missed that. Thanks again Professor!
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 11 Aug 2006, 12:45
Professor's approach is easier:

Men Women
2 (6 men and 2 to choose = 6C2) 1 (4C1)
This one woman being selected cannot be the wife of either of the two men selected earlier. Therefore that leaves you with 4 women to choose from (discarding 2 women who are wives of the two men selected)
= Total 6C2 * 4C1 = 60

1 (4C1)similar logic to above 2 (6C2)
= Total 6C2 * 4C1 = 60

3 (All three selected are men) 0
= Total 6C3 = 20

0 3 (All 3 selected are women)= Total 6C3 = 20

Add 60+60+20+20 = 160
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 17 Mar 2015, 10:08
Just stumbled across this from a google search, so I'll add my technique.

first step: you can only choose one person from each couple, so I found how many combinations of couples I can have:

6c3=20

So there are 20 different ways to combine 3 of the 6 couples.

second step: once I have chosen the 3 couples to combine I have to choose man/wife from each. There are 2 choices in each of the 3 couples so I think of it as a binary number where 0=female and 1=male:

000 = woman,woman,woman
001 = woman,woman,man
...
111 = man,man,man

There are 8 combinations of man/woman from each of the 3 couples (2^3=8)

third step: multiply the number of couple combinations (6c3=20) times the number of man/woman combinations (2^3=8) == 160.

Voila!
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 17 Mar 2015, 10:20
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mand-y wrote:
In how many ways to choose a group of 3 people from 6 couples so that no couple is chosen


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Hope it helps.
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 17 Mar 2015, 20:48
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mand-y wrote:
In how many ways to choose a group of 3 people from 6 couples so that no couple is chosen


Another single step calculation method is that you can select them using the simple basic counting principle (which arranges them in 1st person, 2nd person and 3rd person) and then you can divide by 3! to un-arrange.

First person can be selected in 12 ways.

Second person in 10 ways (since the first person selected and his/her partner are not available)

Third person in 8 ways (since the first and second people and their partners are not available)

Total number of ways = 12*10*8/3! = 160
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 14 Jul 2015, 02:06
VeritasPrepKarishma wrote:
mand-y wrote:
In how many ways to choose a group of 3 people from 6 couples so that no couple is chosen


Another single step calculation method is that you can select them using the simple basic counting principle (which arranges them in 1st person, 2nd person and 3rd person) and then you can divide by 3! to un-arrange.

First person can be selected in 12 ways.

Second person in 10 ways (since the first person selected and his/her partner are not available)

Third person in 8 ways (since the first and second people and their partners are not available)

Total number of ways = 12*10*8/3! = 160


Why do you divide it by 3!? what do you mean by un-arrange?

Thanks
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink] New post 14 Jul 2015, 02:47
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In how many ways to choose a group of 3 people from 6 couples so that no couple is chosen

Another single step calculation method is that you can select them using the simple basic counting principle (which arranges them in 1st person, 2nd person and 3rd person) and then you can divide by 3! to un-arrange.

First person can be selected in 12 ways.

Second person in 10 ways (since the first person selected and his/her partner are not available)

Third person in 8 ways (since the first and second people and their partners are not available)

Total number of ways = 12*10*8/3! = 160

Why do you divide it by 3!? what do you mean by un-arrange?

Thanks


When we use the method of placement of objects like 12*10*8 then we MUST take the cognizance of the arrangement of the objects already included in the method.
e.g. 5*4*3 = 5C3*3!
e.g. 10*9*8 = 10C3*3!



Similarly, 12*10*8 = Selected of three individuals such that no two form couple INCLUDING Arrangement of those three Individuals

therefore, we have to exclude the arrangements of those three individuals (3!) because the question only demand the no. of possible groups and not the arrangement of group members as well.

So only selection of those three individuals so that they are not couple = 12*10*8 / 3! = 160

I hope it helps!
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Re: In how many ways to choose a group of 3 people from 6 couples so that   [#permalink] 14 Jul 2015, 02:47
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