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by eaakbari » Wed Apr 14, 2010 2:44 am
In how many different ways can you choose groups of 3 among 4 married couples so that no husband and wife should be in one group

A 32
B 56
C 24
D NONE


Source : GRE Prep material
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by ajith » Wed Apr 14, 2010 2:53 am
eaakbari wrote:In how many different ways can you choose groups of 3 among 4 married couples so that no husband and wife should be in one group

A 32
B 56
C 24
D NONE


Source : GRE Prep material
Total no of ways to choose 3 people from 8 is 8C3 = 8*7*6/1*2*3 = 56
Total no of ways in which this includes a husband and wife = 4*6 = 24
[one can select the couple in 4 ways and the the third one in the group can be selected in 6 ways from remaining 6]

No of ways in which one can choose so that no husband and wife are in the group = 56-24 = 32
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by eaakbari » Wed Apr 14, 2010 2:56 am
A friend and I had some contention here.

I solved using

8*6*4 = 192

as first place has 8 options and second 6 options and third has 4 options


He solved using

MWM = 4*3*2
WMW=4*3*2
WWW=4C3
MMM=4C3

Summing all you get 56


Someone please tell me why we get this difference as I feel both methods seem correct
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by eaakbari » Wed Apr 14, 2010 3:00 am
Wow ajith, now your method seems to make the most sense out of all :(
Can you tell me what is wrong with my method and my friends too.
Unfortunately I do not have the OA.
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by ajith » Wed Apr 14, 2010 4:01 am
eaakbari wrote:A friend and I had some contention here.

I solved using

8*6*4 = 192

as first place has 8 options and second 6 options and third has 4 options


He solved using

MWM = 4*3*2
WMW=4*3*2
WWW=4C3
MMM=4C3

Summing all you get 56


Someone please tell me why we get this difference as I feel both methods seem correct
Second method is correct unfortunately it includes combinations where Husband and Wife is also there.

First method has a problem with the combination

Say there were four couples A1,A2 B1,B2 C1,C3 and D1,D2

Say you selected A1 first B1 next and D2 next in one
It repeats when B1 is selected first A1 next and D2 last

Basically first method has duplicates.
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by akhpad » Wed Apr 14, 2010 6:40 am
ajith wrote:
eaakbari wrote:In how many different ways can you choose groups of 3 among 4 married couples so that no husband and wife should be in one group

A 32
B 56
C 24
D NONE


Source : GRE Prep material
Total no of ways to choose 3 people from 8 is 8C3 = 8*7*6/1*2*3 = 56
Total no of ways in which this includes a husband and wife = 4*6 = 24
[one can select the couple in 4 ways and the the third one in the group can be selected in 6 ways from remaining 6]

No of ways in which one can choose so that no husband and wife are in the group = 56-24 = 32
I have this problem in my note book and the OA is 32
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by eaakbari » Wed Apr 14, 2010 10:25 am
ajith wrote:
eaakbari wrote:A friend and I had some contention here.

I solved using

8*6*4 = 192

as first place has 8 options and second 6 options and third has 4 options


He solved using

MWM = 4*3*2
WMW=4*3*2
WWW=4C3
MMM=4C3

Summing all you get 56


Someone please tell me why we get this difference as I feel both methods seem correct
Second method is correct unfortunately it includes combinations where Husband and Wife is also there.

First method has a problem with the combination

Say there were four couples A1,A2 B1,B2 C1,C3 and D1,D2

Say you selected A1 first B1 next and D2 next in one
It repeats when B1 is selected first A1 next and D2 last

Basically first method has duplicates.
I understand what you are saying, can you correct the second approach and give the solution
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by Stuart@KaplanGMAT » Wed Apr 14, 2010 12:23 pm
eaakbari wrote:A friend and I had some contention here.

I solved using

8*6*4 = 192

as first place has 8 options and second 6 options and third has 4 options


He solved using

MWM = 4*3*2
WMW=4*3*2
WWW=4C3
MMM=4C3

Summing all you get 56


Someone please tell me why we get this difference as I feel both methods seem correct
As ajith mentioned, the problem with your method is that it includes duplicates.

In your method, you could pick A first, then C second, then F third.

You could also pick C first, A second and F third.

These are identical groups, but you've counted them as different possibilities.

To get from your calculation to the correct answer, you need to eliminate the duplications.

Since you have 3 objects, there are 3! different ways to arrange them. So, you need to divide your answer by 3! to eliminate the duplicates:

192/3! = 192/6 = 32
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by dxgamez » Wed Apr 14, 2010 4:19 pm
Since you have 3 objects, there are 3! different ways to arrange them. So, you need to divide your answer by 3! to eliminate the duplicates:

192/3! = 192/6 = 32
Hi Stuart,

Can this method be used to remove all duplicates wrt no of ways? I was confused when the qn stated no husband and wife should be in the same group. I divided my answer with 2!.

Could you pls explain? Thanks!
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by Fiver » Wed Apr 14, 2010 6:48 pm
eaakbari wrote:In how many different ways can you choose groups of 3 among 4 married couples so that no husband and wife should be in one group

A 32
B 56
C 24
D NONE


Source : GRE Prep material
Another way is to choose 3 couples out of the 4 and then choose 1 from each couple.
4C3 * 2^3 = 4*8 = 32.
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by eaakbari » Wed Apr 14, 2010 10:36 pm
Thanks Stuart, thats a great explanation.
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