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# computation problem

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computation problem [#permalink]  24 Aug 2004, 19:54

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?
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I think I have seen this type of question before.
Say 8 persons are A,B,C,D,E,F,G,H.. AB,CD,EF,GH are couples
Calculate by reducing number of ways when couples will be together from number of total ways.
Number of total ways..8C3
Number of ways when couples will be together.. Say A, B are together..in this case number of ways to select 3rd person from remaining 6 is 6C1=6
Similarly when C,D are together..number of ways to selct 3rd person is 6.
When E, F are together..another 6 ways.
G,H are together..another 6 ways..
So total 24 ways.. when couples are together.
Hence required number of ways = 8C3 -24
=56-24
= 32
So 32 should be the answer
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10C3 for total number of 3 people groups: 120

Consider the couples as (H1,W1), (H2,W2), (H3,W3), (H4,W4), (H5,W5)
For each pair of couple they can pair up with either the husband or wife of another pair to form a group of 3
So for example, H1,W1,H2, H1,W1,W2..etc. So one couple can have total of 8 possible pairings
5 couples will give 5*8=40 pairs

So total number of 3 people group with husband and wife not in the saem group = 120-40 = 80
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Re: computation problem [#permalink]  24 Aug 2004, 23:46
I think it is 80.

10 members. So, 10C3 = 120 is the number of groups formed. This 120 may contain even the couples.

Groups that contain couples only = 5C1 * 8C1 = 40

So, the groups that donot contain couples are 120-40 = 80

lastochka wrote:

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

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Chandra

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I misread the question. There are 10 people and 5 married couples. instead it should be 10C3- 40 =80
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Two ways to think about it:

1) Total - unwanted.
10C3 - 5*8 = 80
5 is for choosing one married couple of the 5, and 8 is for choosing another person out of the remaining8 .

2) 10*8*6/3! = 80.
For the first person you have 10 options.
For the second, only 8.
For the third, only 6.
Divide by 3! because the order doesn't matter.

Hope this helps.
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mallelac, do you mind explaining how you came to the formula for the undesireable outcomes? Wouldn't 5C1+8C1 only give you the number of people that would comprise a 2 person team?
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I am with 80

what is the OA .
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I donot mind at all!!

The undesirable outcome should have a couple. We can choose one out of 5 couples in 5C1 ways thus filling two spots(not one spot). Now, 10-2 = 8 members can be fill up the third spot in 8C1 ways. Thus total ways of undesired outcomes = 5C1*8C1 = 40.

Do let me know if I am not clear.

abennett wrote:
mallelac, do you mind explaining how you came to the formula for the undesireable outcomes? Wouldn't 5C1+8C1 only give you the number of people that would comprise a 2 person team?

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Awaiting response,

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Chandra

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Think of this in terms of a couple being a single unit.

You have 5 couple units.

5C3

now think for each element you've chosen, you could pick husband or wife,
so multiply by 2 for each choice

10 * 2 * 2 * 2 = 80

25s.
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