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Practice Problem 2. from Combinations Lesson

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Practice Problem 2. from Combinations Lesson [#permalink] New post 17 Jul 2007, 07:41
Can someone explain me the answer to this qestion:

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? (Ans. 80)

I thought it is: (10!/(3!*7!)-(5!/(3!*2!))= 110.

Thanks!

Last edited by nalle-b on 17 Jul 2007, 08:36, edited 1 time in total.
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 [#permalink] New post 17 Jul 2007, 08:03
My approach:

Total combinations of choosing 3 out of 10 = 10C3 = 120

Now calculate the number of combinations, which have husband and wife together,
Such combinations will be :

H1W1X - 8 ways

Hence for other couples

So total such ways = 5*8 = 40

Hence the number of combinations, which do not have husband and wife together = 120-40 =80
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 [#permalink] New post 17 Jul 2007, 23:46
Sorry, but I do not understand this step:

Now calculate the number of combinations, which have husband and wife together,
Such combinations will be :

H1W1X - 8 ways

For me it is:

H/W/X
H/X/W
W/H/X
W/X/H
X/H/W
X/W/H

so, I can only see 6 ways.
(Probably it is so easy, but I can´t see it.)
Thanks a lot!
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 [#permalink] New post 18 Jul 2007, 04:24
Hi nalle-b

May be this helps:

we need:
5C1 - combination of choosing 1 pair from 5
8C1 - combination of choosing 1 person from rest 8

5C1 * 8C1 = 40
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 [#permalink] New post 18 Jul 2007, 08:12
Hi,
thanks a lot. I think I got it now.
:idea:
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 [#permalink] New post 18 Jul 2007, 15:52
10*8*6/1*2*3 = 80 (divide by 3 as order is not needed)
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Re: Practice Problem 2. from Combinations Lesson [#permalink] New post 18 Jul 2007, 18:25
nalle-b wrote:
Can someone explain me the answer to this qestion:

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? (Ans. 80)

I thought it is: (10!/(3!*7!)-(5!/(3!*2!))= 110.

Thanks!


2*2*2*(5C3)=8*((5*4)/(1*2))=80
Re: Practice Problem 2. from Combinations Lesson   [#permalink] 18 Jul 2007, 18:25
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Practice Problem 2. from Combinations Lesson

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