in how many ways to choose a group of 3 people from 6

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in how many ways to choose a group of 3 people from 6 [#permalink]

10 Aug 2006, 19:14

in how many ways to choose a group of 3 people from 6 couples so that no couple is chosen

PLZ HELP AND EXPLANATION WANTED

THANKS
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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: PERMUTATIONS [#permalink]

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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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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10 Aug 2006, 23:23

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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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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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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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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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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11 Aug 2006, 13:23

Wow, I cannot believe I missed that. Thanks again Professor!

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Another way to solve this [#permalink]

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

Add 60+60+20+20 = 160
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Business Analyst (Philly, PA)

Another way to solve this [#permalink] 11 Aug 2006, 13:45

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