Ben and Shari can be paired with any of the other three couples, leaving the remaining two for the other team. So, the number of ways to split 4 couples into two teams is 3:
{1, 2} - {3, 4}
{1, 3} - {2, 4}
{1, 4} - {2, 3}

The number of teams where there are no couples:
{Ben, any from couple 2, any from couple 3, any from couple 4} and the remaining 4 people form team #2. The number of team for Ben = 2*2*2 = 8.

Total = 3 + 8 = 11.

Answer: A.

Hope it's clear.
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Hi.. Please check the following solution for this question. I got the correct answer but I want to be sure that I have applied the right method.

So, in this question there are 4 couples i.e a total of 8 people, who need to be divided into 2 teams of 4 member each.

Case 1: when there are 2 couples in each team:
No. of ways = 1/2 (4C2 * 2C2) {I used the 1/2 as the order doesn't matter i.e. Team 1 and Team 2 are interchangeable}

Case 2: No couple in any team
No. of ways = (1/2)*[8*6*4*2][/4*3*2*1] = 8

Therefore, the total no. of possible ways = 8+3 =11

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Yes, that's correct.
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@Bunuel:

Thanks a lot!

hi could you explain this in detail please

Bunuel: Would it be possible for you to kindly explain Case#2 used in the above method.
Case 2: No couple in any team
No. of ways = (1/2)*[8*6*4*2][/4*3*2*1] = 8

Split the two teams into team 1 and team 2. It is easier to define logic in that case. But since we don't actually have distinct team 1 and team 2,
We will divide the answer by 2 to un-arrange the two teams.

n(Two teams with two couples each) -
Select 2 couples for team 1 in 4C2 ways. Team 2 will get the leftover 2 couples.
n (Two non distinct teams with two couples each) = 4C2 /2 = 3 ways

n(Two teams with no couples) -
For team 1, choose the first player in 8 ways, second player in 6 ways, third player in 4 ways and the fourth player in 2 ways. So you get team 1 in 8*6*4*2 ways but players are arranged here. Un-arrange them by dividing by 4!. You get 8*6*4*2/4! for team 1. Leftover 4 people will be in team 2.

n (Two non distinct teams with no couples) = (8*6*4*2/4!) / 2 = 8 ways

Total number of ways = 3 + 8 = 11 ways
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How do you decide by what number to divide for unarranging?
i.e. how did you manage to decide to divide by 4 ! in above case

I really like your answer, can you help me by explaining some more, so if there are 3 teams with 2 couples in each tem, it should be (6C2 * 4C2 * 2C2) / 3!, am i correct?

4 people are arranged - first player, second player, third player and fourth player. 4 people are arranged by multiplying by 4! so if you want to un-arrange them, you will divide by 4!
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Couples can be
AB CD
AC BD
AD BC

No couples can be
Ben M2M3M4 & Shari F2F3F4 (1 combination)
Ben F2M3M4 & Shari M2F3F4 (3 combinations possible with F3 & F4 moving to other side)
Ben F2F3M4 & Shari M2M3F4 (3 combinations possible F2F3, F2F4, F3F4)
Ben F2F3F4 & Shari M2M3M4 (1 combination)

Total 3+3+3+1+1 = 11 combinations

Similar questions to practice?

Hi Bunuel, for the sake of understanding, I listed out all the possible ways of team formation. However I just got 8 ways. What am I doing wrong? Please help.

Posted from my mobile device

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!
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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,D
A,C(Husband) and B,D(Wife) - Team 1
A,C(Wife) and B,D(Husband) - Team 2
A,B1,C1,D and A1,B,C,D1
A,D(Husband) and B,C(Wife) - Team 1
A,D(Wife) and B,C(Husband) - Team 2
A1,B,C,D1 and A,B1,C1,D
B,C(Husband) and A,D(Wife) - Team 1
B,C(Wife) and A,D(Husband) - Team 2

Each of these combinations are unique. Hope this clears out your confusion!
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