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There are 5 married couples and a group of three is to be

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There are 5 married couples and a group of three is to be  [#permalink]

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New post 19 Jul 2008, 16:53
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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?

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html
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Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

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New post 19 Jul 2008, 17:11
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Total ways of selecting 3 people out of 10 = 10C3 = 10*9*8/3*2 = 120

So final answer cannot be greater than 120.

In 3 people only 1 couple is possible. So way of selecting 1 couple out of 5 = 5C1 = 5
Way of selecting 1 person out of remaining 8 = 8C1 = 8

Total ways when there will be 1 couple and 1 different person = 8*5 = 40

Total number of ways in which no people will be from same couple = 120 - 40 = 80
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Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

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New post 19 Jul 2008, 17:17
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I like your approach. When I solved it, I did it a bit differently.

There are 3 spots

Spot #1 can be any of the 10, because no one has been selected so there is no one to avoid.

Spot #2 can choose 1 of only 8 people. 9 people left, but we cannot select the spouse of #1

Spot #3 can choose 1 of only 6 people. 8 people left but we can't select the spouse of #1 or spouse of #2.

Now, we have 8 x 10 x 6 = 480 total possibilities, but we're not done.

This gives us the Permutation of the event, where we have identified order of #1, #2, #3, etc but the order doesn't matter. So we need to divide out the # of times (or permutations) of when we have the same people but in a different order. The total # of perms of a group of 3 is 3x2 = 6

If you remember we had 80 * 6, and we're dividing out 6, this leaves 80.
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Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

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New post 19 Jul 2008, 20:21
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another method : calculate probablity and multiply it by total number of ways = 120

probablity of selecting first person = 10/10 (can be anyone)
probability of selecting secd person = 8/9 ( there are 9 left, but we can not take spouse of 1)
probability of selecting third person = 6/8 (there are 8 left but we can take only 6)

P = 10/10 * 8/9 * 6/8 = 2/3

fav ways = P * total ways = 2/3 * 120 = 80 Answer
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Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

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New post 20 Jul 2008, 07:13
jallenmorris 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?


5 couples:
H1W1 H2W2 H3W3 H4W4 H5W5

group of 3 out of 10 people = 10C3 = 10*9*8/6 = 120

now exclude all the possible arrangements when a Husband and a wife can be in a same group.

Possible such groups:
H1W1H2 H1W1W2 ... H5W5H4 H5W5W4[Thre are 5 pairs and each pair can join with 8 other people to make a BAD husband-wife combinatioin]

5*8=40
12-40 = 80
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Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

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New post 20 Jul 2008, 08:29
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choose 3 couples out of 5, then there are 2 options for each couple (husband or wife)
C(5,3)*2^3=(5*4/2)*8=10*8 = 80
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Re: There are 5 married couples and a group of three is to be  [#permalink]

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New post 27 Jul 2013, 14:01
jallenmorris 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?


what if the questions changed to something like "there are 5 married couples, how many arrangements are there if a husband and wife must stand next to each other?
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Re: There are 5 married couples and a group of three is to be  [#permalink]

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New post 27 Jul 2013, 14:22
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catty2004 wrote:
jallenmorris 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?


what if the questions changed to something like "there are 5 married couples, how many arrangements are there if a husband and wife must stand next to each other?


Then to find the total number I would consider each couple as a single entity, so we have 5 single units that can be arranged in 5! ways.
\(A,B,C,D,E\) or \(A,B,C,E,D\) and so on ...

But within each single unit you can change to position of the husband and the wife \((H,W)\) or \((W,H)\), so within each couple you have 2 possible arrangements.

\(Tot=5!*2^5\)
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Re: There are 5 married couples and a group of three is to be  [#permalink]

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New post 28 Jul 2013, 04:17
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Re: There are 5 married couples and a group of three is to be  [#permalink]

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