Romannepal
pushpitkc
You seem to have missed the following three combinations(in no couples in each team)
A,B1,C,D1 and A1,B,C1,D
A,B1,C1,D and A1,B,C,D1
A1,B,C,D1 and A,B1,C1,D
Also the listed 4th combination(in this section) is A,B1,C,D and A1,B,C1,D1
That takes the grand total to 3(Only couples) + 8(No couples) including the one's I have listed = 11
Hope it helps!
Thanks. However, you have just interchanged the options. Doesn't that mean the same team pair anyway? Please clarify.
I have not interchanged the combinations,
You had listed the four couple to be (A,A1),(B,B1),(C,C1) and (D,D1)
Here A/B/C/D-Husbands and A1,B1,C1,D1-Wives
You had already listed out 5 combinations(for no couples in a team)
Team 1 --------- Team 2
A,B,C,D and A1,B1,C1,D1
A1,B1,C1,D and A,B,C,D1
A1,B1,C,D1 and A,B,C1,D
A1,B,C1,D1 and A,B1,C,D
A,B1,C1,D1 and A1,B,C,D
In addition i had listed the combinations
A,B1,C,D1 and A1,B,C1,DA,C(Husband) and B,D(Wife) - Team 1
A,C(Wife) and B,D(Husband) - Team 2A,B1,C1,D and A1,B,C,D1A,D(Husband) and B,C(Wife) - Team 1
A,D(Wife) and B,C(Husband) - Team 2A1,B,C,D1 and A,B1,C1,DB,C(Husband) and A,D(Wife) - Team 1
B,C(Wife) and A,D(Husband) - Team 2Each of these combinations are unique. Hope this clears out your confusion!