Total no of ways to choose 3 people from 8 is 8C3 = 8*7*6/1*2*3 = 56eaakbari 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
Second method is correct unfortunately it includes combinations where Husband and Wife is also there.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
I have this problem in my note book and the OA is 32ajith wrote:Total no of ways to choose 3 people from 8 is 8C3 = 8*7*6/1*2*3 = 56eaakbari 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 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 understand what you are saying, can you correct the second approach and give the solutionajith wrote:Second method is correct unfortunately it includes combinations where Husband and Wife is also there.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
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.
As ajith mentioned, the problem with your method is that it includes duplicates.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
Hi Stuart,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
Another way is to choose 3 couples out of the 4 and then choose 1 from each couple.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
