Ben and Shari have invited three other couples to their apar

Question

Ben and Shari have invited three other couples to their apartment for a game of charades. Everyone is to be split into two equal teams, but each team must consist of either two couples or no couples. How many different pairs of teams are possible?

(A) 11
(B) 14
(C) 22
(C) 35
(D) 70

(A) 11
(B) 14
(C) 22
(C) 35
(D) 70

Bunuel · Accepted Answer

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.

KarishmaB · Answer

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

antra.nith@gmail.com · Answer

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

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

Bunuel · Answer

____________
Yes, that's correct.

antra.nith@gmail.com · Answer

@Bunuel:

Thanks a lot!

yaash · Answer

hi could you explain this in detail please

sameerspice · Answer

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

KM2018AA · Answer

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

chesstitans · Answer

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?

KarishmaB · Answer

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!

sriramkrishnan · Answer

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

ShashankDave · Answer

Similar questions to practice?

Romannepal · Answer

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

pushpitkc · Answer

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!

Romannepal · Answer

Thanks. However, you have just interchanged the options. Doesn't that mean the same team pair anyway? Please clarify.

pushpitkc · Answer

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!

Regor60 · Answer

Call the 4 couples

ABCD

Once you pair up 2 couples to form a team the other team is automatically defined:

AB CD
AC BD
AD BC

So 3 ways

For the teams of non couples,
the first person can be selected 8 ways, the second 6 ways, since the other person from the first couple can't be selected,and 4 and 2 for the remainder teammates following the same logic:

8 6 4 2
_ _ _ _

The remaining 4 people form the other team.

So 8*6*4*2. However, need to divide by 4*3*2*1 since the 4 are unnecessarily permuted.

Also need to divide by 2 because the two teams are repeated with this approach, so

(8*6*4*2)/(4*3*2*2) = 8

Total ways = 3+8= 11

Posted from my mobile device

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Ben and Shari have invited three other couples to their apar

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