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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] New post 30 Jan 2004, 17:19
* 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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 [#permalink] New post 30 Jan 2004, 17:25
C(10,3) = 120 (with no restrictions)

# ways to have a couple in the group of 3 = 5 * 8 = 40

Answer, I guess, is = 80
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 [#permalink] New post 30 Jan 2004, 17:45
Hmmmmmm I am getting 560.
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 [#permalink] New post 30 Jan 2004, 19:42
Hi, I am coming over from the 'other' test prep site. Found this question over there as well.

I agree with Zhung Gazi on this one.

10c3 yields 120 with no restrictions.

figuring how many groups can be formed with couples being apart of them, would be 5 * 8 or 40.

120 - 40 yields 80 distinct groups.
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 [#permalink] New post 04 Feb 2004, 03:25
Total Combinations :- 10C3=120
Ways in which one couple can be involved in the group is = 8
so for all couples it comes out to be = 5*8= 40

So total ways in which no couple is involved in the group = 120-40 = 80
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 [#permalink] New post 04 Feb 2004, 09:44
I guess the way to solve this would be

total-# of ways a husband and wife can be in a group

total - 10C3 = 120

Ways in which couple can be a part of the group 5*8 = 40

5 Couples we can choose 1 = 5C1 = 5
1 Person from the remaining 8 = 8C1 = 8

Ans : 120-40.

Please correct me If I am wrong
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 [#permalink] New post 04 Feb 2004, 20:37
Pakoo u r right...

Vivek.
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  [#permalink] 04 Feb 2004, 20:37
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