Romannepal wrote:
pushpitkc wrote:
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!
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.
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