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

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SVP
Joined: 30 Apr 2008
Posts: 1830
Location: Oklahoma City
Schools: Hard Knocks
There are 5 married couples and a group of three is to be  [#permalink]

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19 Jul 2008, 16:53
2
00:00

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(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 6 sessions

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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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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Director Joined: 10 Sep 2007 Posts: 887 Re: PS:Comb/Perm Married Couples...not swingers ;-) [#permalink] ### Show Tags 19 Jul 2008, 17:11 4 2 Total ways of selecting 3 people out of 10 = 10C3 = 10*9*8/3*2 = 120 So final answer cannot be greater than 120. In 3 people only 1 couple is possible. So way of selecting 1 couple out of 5 = 5C1 = 5 Way of selecting 1 person out of remaining 8 = 8C1 = 8 Total ways when there will be 1 couple and 1 different person = 8*5 = 40 Total number of ways in which no people will be from same couple = 120 - 40 = 80 SVP Joined: 30 Apr 2008 Posts: 1830 Location: Oklahoma City Schools: Hard Knocks Re: PS:Comb/Perm Married Couples...not swingers ;-) [#permalink] ### Show Tags 19 Jul 2008, 17:17 3 I like your approach. When I solved it, I did it a bit differently. There are 3 spots Spot #1 can be any of the 10, because no one has been selected so there is no one to avoid. Spot #2 can choose 1 of only 8 people. 9 people left, but we cannot select the spouse of #1 Spot #3 can choose 1 of only 6 people. 8 people left but we can't select the spouse of #1 or spouse of #2. Now, we have 8 x 10 x 6 = 480 total possibilities, but we're not done. This gives us the Permutation of the event, where we have identified order of #1, #2, #3, etc but the order doesn't matter. So we need to divide out the # of times (or permutations) of when we have the same people but in a different order. The total # of perms of a group of 3 is 3x2 = 6 If you remember we had 80 * 6, and we're dividing out 6, this leaves 80. _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Director
Joined: 27 May 2008
Posts: 527
Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

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19 Jul 2008, 20:21
2
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
Senior Manager
Joined: 07 Jan 2008
Posts: 267
Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

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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
Director
Joined: 01 Jan 2008
Posts: 595
Re: PS:Comb/Perm Married Couples...not swingers ;-)  [#permalink]

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20 Jul 2008, 08:29
2
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
Manager
Joined: 30 May 2008
Posts: 53
Re: There are 5 married couples and a group of three is to be  [#permalink]

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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?
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1073
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Re: There are 5 married couples and a group of three is to be  [#permalink]

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27 Jul 2013, 14:22
1
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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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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01 Nov 2018, 06:59
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Re: There are 5 married couples and a group of three is to be &nbs [#permalink] 01 Nov 2018, 06:59
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