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

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06 Feb 2005, 16:55

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.

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).

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).

...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?
_________________

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.

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.

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.

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.

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.