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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 03 Jun 2007, 07:08
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?

CAn someone show the logic in this one...or the link if its a repost
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 [#permalink] New post 03 Jun 2007, 07:30
80.

-Every woman can make a group with 2 men out of 4 (5-her husband).
There are 5 women so 5*2C4=30

-Every man can make a group with 2 women out of 4 (5-his wife).
There are 5 men so 5*2C4=30

-5 Women make among themselves 3C5=10 groups
-5Men make among themselves 3C5=10 groups
so 30+30+10+10= 80 groups

If it is not true I am sorry. I am not good at Permutation/Combination and Probability.

cheers,

Last edited by UMB on 03 Jun 2007, 08:04, edited 2 times in total.
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 [#permalink] New post 03 Jun 2007, 07:41
40

Total ppl = 10

Total number of ways a group of 3 can be formed = 10C3 = 120

Total number of ways a grp of 3 is formed with couples = 8C1

now, the couple chosen can be any one of the 5 couples and also either the Husband of the Wife = 5 x 2 = 10 combinations

Hence total number of grp of 3 which includes a couple = 10 x 8C1 = 80

Hence desired number of groups with no couples = 120-80 = 40
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 [#permalink] New post 03 Jun 2007, 07:42
nope. OA is different
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 [#permalink] New post 03 Jun 2007, 07:50
I got 80.

There are two ways of reasoning to solve this problem...it depends on which you are more comfortable with.

1) We are choosing 3 people out of 10, so picture three slots. The possibilities for the first slot is 10. The possibilities for the 2nd slot are now only 8 (not 9, because we don't want to choose the wife or husband of the first person). So now we have 10 x 8. For the third slot, we only have 6 people to choose from, once again because we don't want to choose the wives/husbands of the first two people chosen. If you can't picture this, then draw out 8 dots, each pair representing a couple and go through the steps again.

So we have 10x8x6 = 480 possibilities for having a group of people without a married couple. However, this is including duplicates, and since order doesn't matter, we need to divide this by 3!, because we are choosing 3 people.

480/3! = 480/(3x2) = 80.


2.) Using this method, we find the total number of outcomes, which is 10C3 = 120. We now have to find the unfavorable outcomes to subtract from 120. Unfavorable outcomes are where there is a couple in the group of 3. So assuming the first two chosen are a married couple, then for the third person in the group, there are 8 possibilities (the 8 other people not chosen). Since we have 5 married couples, each with 8 possibilities for the 3rd person, we can calculate the number of unfavorable outcomes to be 8x5=40.

120 total possible outcomes - 40 unfavorable outcomes = 80
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 [#permalink] New post 03 Jun 2007, 08:08
plaguerabbit. Thanks
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 [#permalink] New post 03 Jun 2007, 19:56
oh yes .. i multiplied 2 - to consider husband or wife in a given couple .. lol dumb.
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 [#permalink] New post 03 Jun 2007, 20:26
Total arrangements 10*8*6= 480 . Since order does not matter 480/3!=80
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 [#permalink] New post 04 Jun 2007, 04:43
For me is 10*8*6 = 480....what is the OA???

the rationale is that we have for the first round 10 options and them we can take 3 if we do not want a husband or wife together we get back one and we have 8 options , we keep the same rationale and we have 6. finnaly we will have 3 pairs and the probability with no same husb and wife together in the group.
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 [#permalink] New post 04 Jun 2007, 06:28
80

Total groups possible: 10c3 = 120
Total groups that have a couple = 5 * 8c1 = 40
(Start with a couple which there are 5 of, and then pick 1 person from the remaining 8 people)

Total groups without a couple = 120 - 40 = 80
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 [#permalink] New post 04 Jun 2007, 06:33
andrehaui wrote:
For me is 10*8*6 = 480....what is the OA???

the rationale is that we have for the first round 10 options and them we can take 3 if we do not want a husband or wife together we get back one and we have 8 options , we keep the same rationale and we have 6. finnaly we will have 3 pairs and the probability with no same husb and wife together in the group.


This approach doesn't work because you don't count for duplicates (in other words, you are counting the order of people in a group).

What you will have to do to get the right answer that way is to divide your final answer by 6 (which is 3p3 = 6 to account for the duplicates)

So it will be (10 * 8 * 6) / 6 = 80
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 [#permalink] New post 04 Jun 2007, 11:17
ok I see :lol: :oops: :lol: ..but the method of taking out 40 from 120 is better. Thanks. :wink:
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 [#permalink] New post 04 Jun 2007, 17:04
Quote:
Since we have 5 married couples, each with 8 possibilities for the 3rd person, we can calculate the number of unfavorable outcomes to be 8x5=40.
Hi, will someone explain this part. If we have one married couple already chosen there is room for one other person. There are 8 people to choose from. I get that. Now, what was was typed is unclear to me. Can/will someone try and clarify.
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 [#permalink] New post 04 Jun 2007, 18:09
ggarr wrote:
Quote:
Since we have 5 married couples, each with 8 possibilities for the 3rd person, we can calculate the number of unfavorable outcomes to be 8x5=40.
Hi, will someone explain this part. If we have one married couple already chosen there is room for one other person. There are 8 people to choose from. I get that. Now, what was was typed is unclear to me. Can/will someone try and clarify.


I suppose I can try, since I'm the one that wrote it =P

You already understand the gist of it...you just have to realize that each of these 8 scenarios for one married couple is an unfavorable event.

so we have

8 scenarios for couple 1 that are unfavorable

8 scenarios for couple 2 that are unfavorable...

...8 scenarios for couple 5 that are unfavorable.

add them all up, and we get a total of 40 possible scenarios where a couple is chosen.

I hope I clarified it a little bit...good luck
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Re: Marriage!! [#permalink] New post 04 Jun 2007, 18:11
Dek 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?

CAn someone show the logic in this one...or the link if its a repost



Couples

A : a1 a2

B : b1 b2

C : c1 c2

D : d1 d2

E : e1 e2

Total number of people = 10

Number of ways of selecting 3 people out of 10 = 10C3 = 120

Let's find out the number of ways in which a pair of husband and wife are always in the group of three people selcted out of 10.

Couple A ( selected in only 1 way ) X 8C1 ( one person to be selected from the remaining 8 )

Now this can be repeated in 5 different ways because thre are 5 different couples.

for example;

Couple B ( selected in only 1 way ) X 8C1 ( one person to be selected from the remaining 8 ) etc.

thus; 5*8C1 = 40 ways.

(total number of ways of selecting 3 people) - ( number of ways of always selecting a couple ) = number of ways of not selecting the couple

therefore; 120-40 = 80 ways.

Hope this helps!
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 [#permalink] New post 04 Jun 2007, 18:19
Thank you Plaugerabbit and LM
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Re: [#permalink] New post 17 Jul 2007, 02:39
I got till the 480 point.. Could anyone explain what is meant by duplicates for which we are dividing by 3!.

When we consider 10 ppl in the first step and then the remaining 8 in the next step and then 6, how can there be duplicates. The ones considered in evaluating 10 are not being used in evaluating the remaining 8.

Thanks
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 [#permalink] New post 17 Jul 2007, 08:43
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Re: Marriage!! [#permalink] New post 17 Jul 2007, 20:31
Dek 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?

CAn someone show the logic in this one...or the link if its a repost


10/10 x 8/9 x 6/8 = 480/720 = 2/3

2/3 of all arrangements are groups where husband and wife are not in the same group

10C3 = 120
120 X 2/3 = 80
Re: Marriage!!   [#permalink] 17 Jul 2007, 20:31
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