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

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

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03 Jul 2004, 11:51
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There are 5 married couples and a groups of three is to be formed out of them; how many arrangements sre there if a hasband and wife may not be in the same group?

I couldn't reach the OA . can someone help me with that
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Last edited by dr_sabr on 04 Jul 2004, 01:36, edited 1 time in total.

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03 Jul 2004, 12:15
I think the answer is 80.

Total ways to form 3 from 10 (5 couples) = C(10,3) = 120.

Groups with same couples = 8 * 5 = 40
where 5==> no. of couples
8 ==> If a couple is chosen than there are 8 people to choose the last person from.

Ans 120-40 =80.

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03 Jul 2004, 20:15

This seems to be a tricky one. Here's my attempt.

Since we are supposed to make groups of three ,consider

__ __ __ as a group of men and women.

The first slot can be filled by any one of the 10 men or women.

Now 2 people are booked. One is in the group and one is a spouse.
The second slot can be filled by only one of the 8 men or women.
The 9th person is the spouse of the first person, and can't fill this slot.

So now 4 people are booked. 2 are in the group and 2 are spouses.

The third slot can be filled by only one of the 6 men or women.

Total no of combinations = 10*8*6 = 680

I am not sure about my answer. There might be a flaw in it. Let me know if you find any.

- ash
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ash
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03 Jul 2004, 22:51
Ash,

I believe you are under the assumption that order the couples are arranged matters.

But I dont think the order should matter here. We should employ combination principle rather then permutation.

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03 Jul 2004, 22:52
Ash,

I believe you are under the assumption that order the couples are arranged matters.

But I dont think the order should matter here. We should employ combination principle rather then permutation.

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04 Jul 2004, 11:06
I think it's 80 too.

If you would choose them and put them on a row it would be 10*8*6=480.
10 For the first
8 For the second
6 for the third

But this is not a row, so you have to divide by 3!=6.

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04 Jul 2004, 21:46
ashkg wrote:

This seems to be a tricky one. Here's my attempt.

Since we are supposed to make groups of three ,consider

__ __ __ as a group of men and women.

The first slot can be filled by any one of the 10 men or women.

Now 2 people are booked. One is in the group and one is a spouse.
The second slot can be filled by only one of the 8 men or women.
The 9th person is the spouse of the first person, and can't fill this slot.

So now 4 people are booked. 2 are in the group and 2 are spouses.

The third slot can be filled by only one of the 6 men or women.

Total no of combinations = 10*8*6 = 680

I am not sure about my answer. There might be a flaw in it. Let me know if you find any.

- ash

10*8*6= 480 ( i made a mistake there)

another mistake is that I was supposed to make selections and not permutations.

final answer = 480/3! = 80

Thanks sathya !

- ash
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ash
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06 Jul 2004, 02:09
The committe could be made like this (W for wives, H for husbands)

3H : 5C3 = 10 ways
2H + 1W : 5C2*3C1 = 30 ways (excluding wives of selected husbands)
1H + 2W : 5C1*4C2 = 30 ways (--do--)
3W : 5C3 = 10 ways

Total = 80 ways.

Hope it helps
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06 Jul 2004, 10:18
This is quite simple:
5 couples so 10 individuals.
so total arrangements wud be 10c3= 120
if we had to selecta group with a couple it wud be 5c1 for the first 2 spot and 8c1 for the third spot in the group
then 10c3-8c1x5c1
120-8x5= 80

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06 Jul 2004, 10:18
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