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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 06 Feb 2005, 16:55
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 can't seem to get 80. Help appreciated. Thank you.
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 [#permalink] New post 06 Feb 2005, 16:57
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 [#permalink] New post 06 Feb 2005, 17:00
I get 480. Can you confirm the answer and explaination u have?
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 [#permalink] New post 06 Feb 2005, 21:03
C(10,3)-C(5,1)*C(8,1)=120-40=80
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 [#permalink] New post 08 Feb 2005, 21:32
HongHu wrote:
C(10,3)-C(5,1)*C(8,1)=120-40=80


HongHu,

could you please explain this part 5C1 * 8C1?
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 [#permalink] New post 08 Feb 2005, 21:38
You need to use total combination minus the special cases where a couple was picked.

So you pick a couple. Which is C(5,1). Now you still need one more person to form the 3 people committee from the remaing 8 people. There are C(8,1) ways of doing this. In other words the number of combinations where a couple was picked is C(5,1)*C(8,1).
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 [#permalink] New post 08 Feb 2005, 22:01
HongHu wrote:
You need to use total combination minus the special cases where a couple was picked.

So you pick a couple. Which is C(5,1). Now you still need one more person to form the 3 people committee from the remaing 8 people. There are C(8,1) ways of doing this. In other words the number of combinations where a couple was picked is C(5,1)*C(8,1).


Great. Thanks HongHu
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 [#permalink] New post 09 Feb 2005, 12:50
...I was wondering if you could do it this way and would like some input:


AB CD EF GH IJ

represent 5 married couples

- - - <--- represents three spots


If you pick A you have 8 choices between the rest excluding B. This apllies to the other ten members. Therefore the answer is 8*10 = 80 Does this work?
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 [#permalink] New post 09 Feb 2005, 14:19
Hmmm are you only making a 2 people committee?
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 [#permalink] New post 09 Feb 2005, 20:34
Since you need a committee of 3 select 3 couples from the 5, 5 choose 3 = 10 ways. Then there are 2 choices from each couple. Therefore the total arrangements is 10 * (2*2*2) = 80.
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Re: Probability question [#permalink] New post 21 Feb 2005, 14:15
sk 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? (Ans. 80)

I can't seem to get 80. Help appreciated. Thank you.


HongHu's method makes all the sense, however I keep getting stuck with my approach:
10*8*6 = 480

ten ways to pick first person in a trio, 8 ways to pick one out of the remaining 8 people, and 6 ways to pick the last, third person.

Where's the flaw in this logic? Thanks!
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 [#permalink] New post 21 Feb 2005, 15:00
lastochka,
I agree with your question.Your logic seems perfectly fine to me.

As we know the answer backsolving tells us answer can be 10*8*6 /6,that means somehow your formula has 6 times repeats,so you need to devide it by 6.

Now mystery is where the repitions are.
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 [#permalink] New post 21 Feb 2005, 15:17
lastochka,
After a long thought,I guess I figured the problem.

Let's say there is no restriction and wife&husband can be in the same group.Then the problem is down to picking 3 from 10. According to your logic, you will pick 10*9*8. However,it is obiously wrong, order is not important.M1,M2,M3 are same as M1,M3,M2.So it can not be permutations,it should be combinations.

Looking from the fundamental diff from permutations,in combinations, order is not important,so you devide the P by r! and you will get C.
nCr = nPr/r!

Similarly in the original example,since you picked the perumations way, you need to devide it by r! which is 3! .So 10*8*6/3!=80.

Thanks for raising the question thus helping refresh my rusty math concepts.
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 [#permalink] New post 21 Feb 2005, 19:42
700Plus wrote:
lastochka,
After a long thought,I guess I figured the problem.

Let's say there is no restriction and wife&husband can be in the same group.Then the problem is down to picking 3 from 10. According to your logic, you will pick 10*9*8. However,it is obiously wrong, order is not important.M1,M2,M3 are same as M1,M3,M2.So it can not be permutations,it should be combinations.

Looking from the fundamental diff from permutations,in combinations, order is not important,so you devide the P by r! and you will get C.
nCr = nPr/r!

Similarly in the original example,since you picked the perumations way, you need to devide it by r! which is 3! .So 10*8*6/3!=80.

Thanks for raising the question thus helping refresh my rusty math concepts.


appreciate the dicussion here
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 [#permalink] New post 21 Feb 2005, 23:19
10!/3!7! = 120 --> Total number of groups
Number of arrangements husband and wife in the same group:
= H1W1(H2), H1W1(W2).. etc

Each pair can be paired up 8 times
Total of 5 pairs, so 40 combinations where husband and wife are in the same team.

So number of groups where husband and wife are not in the same team
= 120-40 =80
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 [#permalink] New post 22 Feb 2005, 07:39
baggarwal wrote:
Since you need a committee of 3 select 3 couples from the 5, 5 choose 3 = 10 ways. Then there are 2 choices from each couple. Therefore the total arrangements is 10 * (2*2*2) = 80.

Yes, a good way to solve this kind of question.

700Plus wrote:
Looking from the fundamental diff from permutations,in combinations, order is not important,so you devide the P by r! and you will get C.
nCr = nPr/r!

Similarly in the original example,since you picked the perumations way, you need to devide it by r! which is 3! .So 10*8*6/3!=80.

Another great approach!

A lot of times there are multiple approaches to a permutation/combination question. If you have time it may be good that you could use different methods to verify your solution.
  [#permalink] 22 Feb 2005, 07:39
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