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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]

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10 Aug 2006, 19:14
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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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when there is a will there is a way

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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink]

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10 Aug 2006, 19:28
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]

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10 Aug 2006, 20:30
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]

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10 Aug 2006, 22:35
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]

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10 Aug 2006, 23:23
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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]

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11 Aug 2006, 04:21
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]

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11 Aug 2006, 07:08
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]

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11 Aug 2006, 07:17
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]

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11 Aug 2006, 11:27
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]

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11 Aug 2006, 13:23
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]

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11 Aug 2006, 13:45
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

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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink]

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17 Mar 2015, 11:08
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]

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17 Mar 2015, 11:20
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink]

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17 Mar 2015, 21:48
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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]

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14 Jul 2015, 03:06
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]

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14 Jul 2015, 03:47
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patufet6 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

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]

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14 Oct 2015, 05:23
mand-y wrote:
In how many ways to choose a group of 3 people from 6 couples so that no couple is chosen

Can somebody help me out with this?
Multiple doubts!

1. First approach was to select any of the 12 people. As there is another person who will correspond to being a couple there are 10 people and similarly 8 choices for the third place:
12*10*8

2. If I take the other approach and subtract no. of ways of selecting a couple and subtract it from the total ways to select 3, how do we get the number of selecting a couple as 60 as mentioned in some posts above?

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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink]

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14 Oct 2015, 06:29
longfellow wrote:
mand-y wrote:
In how many ways to choose a group of 3 people from 6 couples so that no couple is chosen

Can somebody help me out with this?
Multiple doubts!

1. First approach was to select any of the 12 people. As there is another person who will correspond to being a couple there are 10 people and similarly 8 choices for the third place:
12*10*8

2. If I take the other approach and subtract no. of ways of selecting a couple and subtract it from the total ways to select 3, how do we get the number of selecting a couple as 60 as mentioned in some posts above?

1. First approach was to select any of the 12 people. As there is another person who will correspond to being a couple there are 10 people and similarly 8 choices for the third place:

12*10*8

This method includes the arrangement of those three individuals to be selected. Since we only require Selection so we shall exclude all arrangements of those three individuals by dividing the result by 3! (no. of ways in which 3 individuals can be arranged amongst each other)

i.e. 12*10*8/3! = 160

2. If I take the other approach and subtract no. of ways of selecting a couple and subtract it from the total ways to select 3, how do we get the number of selecting a couple as 60 as mentioned in some posts above?
Total ways to select 3 out of 12 people = 12C3 = 220
No. of ways selecting three people such that 2 of them are a couple = 6C1 * 10 = 60
6C1 = No. of ways of selecting one couple i.e. 2 people out of 6 couples
10 = No. of ways of selecting third person out of remaining 10 individuals (5 couples)
Favorable ways = 220 - 60 = 160

I hope this helps!
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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink]

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14 Oct 2015, 07:04
GMATinsight wrote:
longfellow wrote:
mand-y wrote:
In how many ways to choose a group of 3 people from 6 couples so that no couple is chosen

Can somebody help me out with this?
Multiple doubts!

1. First approach was to select any of the 12 people. As there is another person who will correspond to being a couple there are 10 people and similarly 8 choices for the third place:
12*10*8

2. If I take the other approach and subtract no. of ways of selecting a couple and subtract it from the total ways to select 3, how do we get the number of selecting a couple as 60 as mentioned in some posts above?

1. First approach was to select any of the 12 people. As there is another person who will correspond to being a couple there are 10 people and similarly 8 choices for the third place:

12*10*8

This method includes the arrangement of those three individuals to be selected. Since we only require Selection so we shall exclude all arrangements of those three individuals by dividing the result by 3! (no. of ways in which 3 individuals can be arranged amongst each other)

i.e. 12*10*8/3! = 160

2. If I take the other approach and subtract no. of ways of selecting a couple and subtract it from the total ways to select 3, how do we get the number of selecting a couple as 60 as mentioned in some posts above?
Total ways to select 3 out of 12 people = 12C3 = 220
No. of ways selecting three people such that 2 of them are a couple = 6C1 * 10 = 60
6C1 = No. of ways of selecting one couple i.e. 2 people out of 6 couples
10 = No. of ways of selecting third person out of remaining 10 individuals (5 couples)
Favorable ways = 220 - 60 = 160

I hope this helps!

Thanks for the quick response. Could you please elaborate on this concept of dividing arrangements by factorial to get selections as I can't get my head around it. As per my opinion, I simply applied the counting method to get the answer.

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Re: In how many ways to choose a group of 3 people from 6 couples so that [#permalink]

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14 Oct 2015, 08:56
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longfellow wrote:

Thanks for the quick response. Could you please elaborate on this concept of dividing arrangements by factorial to get selections as I can't get my head around it. As per my opinion, I simply applied the counting method to get the answer.

If you have to select 3 individuals out of 5 people then you solve it like 5C3 = 10

But when we solve the question line filling three blank spaces by those 5 individuals then the method says

5 options for first place
4 options for Second place
3 options for Third place
i.e. total ways to fill three spaces with 3 of 5 people = 5*4*3 = 60
But this 60 includes the arrangement of those 3 individuals i.e. One of the arrangements of three people A, B and C 'ABC' will be treated different from arrangement 'ACB'

Now, This same question can be answered as 5C3*3! = 10*6 = 60

i.e. 5*4*3 = 5C3*3!
i.e. 5C3 = 5*4*3/3! Which is pure selection of 3 out of 5 individuals.

I hope this helps!
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Re: In how many ways to choose a group of 3 people from 6 couples so that   [#permalink] 14 Oct 2015, 08:56

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