It is currently 18 Oct 2017, 02:38

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Intern
Joined: 01 Nov 2003
Posts: 1

Kudos [?]: [0], given: 0

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

### Show Tags

06 Feb 2005, 17:55
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.

Kudos [?]: [0], given: 0

VP
Joined: 18 Nov 2004
Posts: 1431

Kudos [?]: 44 [0], given: 0

### Show Tags

06 Feb 2005, 17:57

Kudos [?]: 44 [0], given: 0

Manager
Joined: 01 Jan 2005
Posts: 165

Kudos [?]: 4 [0], given: 0

Location: NJ

### Show Tags

06 Feb 2005, 18:00
I get 480. Can you confirm the answer and explaination u have?

Kudos [?]: 4 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2231

Kudos [?]: 376 [0], given: 0

### Show Tags

06 Feb 2005, 22:03
C(10,3)-C(5,1)*C(8,1)=120-40=80

Kudos [?]: 376 [0], given: 0

Director
Joined: 27 Dec 2004
Posts: 896

Kudos [?]: 53 [0], given: 0

### Show Tags

08 Feb 2005, 22:32
HongHu wrote:
C(10,3)-C(5,1)*C(8,1)=120-40=80

HongHu,

could you please explain this part 5C1 * 8C1?

Kudos [?]: 53 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2231

Kudos [?]: 376 [0], given: 0

### Show Tags

08 Feb 2005, 22:38
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).

Kudos [?]: 376 [0], given: 0

Director
Joined: 27 Dec 2004
Posts: 896

Kudos [?]: 53 [0], given: 0

### Show Tags

08 Feb 2005, 23:01
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

Kudos [?]: 53 [0], given: 0

Senior Manager
Joined: 30 Dec 2004
Posts: 294

Kudos [?]: 3 [0], given: 0

Location: California

### Show Tags

09 Feb 2005, 13:50
...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?
_________________

"No! Try not. Do. Or do not. There is no try.

Kudos [?]: 3 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2231

Kudos [?]: 376 [0], given: 0

### Show Tags

09 Feb 2005, 15:19
Hmmm are you only making a 2 people committee?

Kudos [?]: 376 [0], given: 0

Intern
Joined: 06 Jan 2005
Posts: 23

Kudos [?]: [0], given: 0

### Show Tags

09 Feb 2005, 21:34
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.

Kudos [?]: [0], given: 0

Senior Manager
Joined: 07 Oct 2003
Posts: 349

Kudos [?]: 21 [0], given: 0

Location: Manhattan

### Show Tags

21 Feb 2005, 15:15
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!

Kudos [?]: 21 [0], given: 0

Manager
Joined: 27 Jan 2005
Posts: 100

Kudos [?]: 1 [0], given: 0

Location: San Jose,USA- India

### Show Tags

21 Feb 2005, 16:00
lastochka,

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.

Kudos [?]: 1 [0], given: 0

Manager
Joined: 27 Jan 2005
Posts: 100

Kudos [?]: 1 [0], given: 0

Location: San Jose,USA- India

### Show Tags

21 Feb 2005, 16:17
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.

Kudos [?]: 1 [0], given: 0

Senior Manager
Joined: 07 Oct 2003
Posts: 349

Kudos [?]: 21 [0], given: 0

Location: Manhattan

### Show Tags

21 Feb 2005, 20:42
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

Kudos [?]: 21 [0], given: 0

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5036

Kudos [?]: 437 [0], given: 0

Location: Singapore

### Show Tags

22 Feb 2005, 00:19
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

Kudos [?]: 437 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2231

Kudos [?]: 376 [0], given: 0

### Show Tags

22 Feb 2005, 08:39
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.

Kudos [?]: 376 [0], given: 0

22 Feb 2005, 08:39
Display posts from previous: Sort by