Last visit was: 15 Jun 2025, 11:24 It is currently 15 Jun 2025, 11:24
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
avatar
Ronak275
Joined: 16 Dec 2013
Last visit: 26 Aug 2014
Posts: 9
Own Kudos:
81
 [54]
Given Kudos: 5
Posts: 9
Kudos: 81
 [54]
4
Kudos
Add Kudos
50
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 15 June 2025
Posts: 102,025
Own Kudos:
Given Kudos: 93,857
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,025
Kudos: 732,541
 [19]
5
Kudos
Add Kudos
14
Bookmarks
Bookmark this Post
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 15 Jun 2025
Posts: 16,046
Own Kudos:
73,682
 [7]
Given Kudos: 472
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,046
Kudos: 73,682
 [7]
2
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
General Discussion
avatar
[email protected]
Joined: 18 Jan 2014
Last visit: 24 May 2015
Posts: 4
Own Kudos:
8
 [3]
Given Kudos: 4
Location: United States
Concentration: General Management, Operations
GMAT 1: 760 Q51 V42
GMAT 1: 760 Q51 V42
Posts: 4
Kudos: 8
 [3]
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 15 June 2025
Posts: 102,025
Own Kudos:
Given Kudos: 93,857
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,025
Kudos: 732,541
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
____________
Yes, that's correct.
avatar
[email protected]
Joined: 18 Jan 2014
Last visit: 24 May 2015
Posts: 4
Own Kudos:
Given Kudos: 4
Location: United States
Concentration: General Management, Operations
GMAT 1: 760 Q51 V42
GMAT 1: 760 Q51 V42
Posts: 4
Kudos: 8
Kudos
Add Kudos
Bookmarks
Bookmark this Post
@Bunuel:

Thanks a lot!
avatar
yaash
Joined: 22 Nov 2014
Last visit: 29 Mar 2017
Posts: 22
Own Kudos:
4
 [1]
Given Kudos: 160
Posts: 22
Kudos: 4
 [1]
Kudos
Add Kudos
Bookmarks
Bookmark this Post
bumpbot
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.


hi could you explain this in detail please
avatar
sameerspice
Joined: 16 Apr 2015
Last visit: 31 Jul 2020
Posts: 24
Own Kudos:
Given Kudos: 132
Posts: 24
Kudos: 126
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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
____________
Yes, that's correct.

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
avatar
KM2018AA
Joined: 11 Sep 2016
Last visit: 08 Feb 2021
Posts: 63
Own Kudos:
Given Kudos: 23
Location: India
Concentration: General Management, Leadership
GMAT 1: 710 Q47 V40
GPA: 3
WE:Sales (Manufacturing)
Products:
GMAT 1: 710 Q47 V40
Posts: 63
Kudos: 21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma
Ronak275
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

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

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
User avatar
chesstitans
Joined: 12 Dec 2016
Last visit: 20 Nov 2019
Posts: 991
Own Kudos:
Given Kudos: 2,562
Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
GMAT 1: 700 Q49 V33
Posts: 991
Kudos: 1,892
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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

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?
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 15 Jun 2025
Posts: 16,046
Own Kudos:
73,682
 [2]
Given Kudos: 472
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,046
Kudos: 73,682
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
KM2018AA
VeritasPrepKarishma
Ronak275
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

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

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

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!
User avatar
sriramkrishnan
Joined: 03 Jan 2016
Last visit: 18 May 2021
Posts: 15
Own Kudos:
Given Kudos: 6
Posts: 15
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
ShashankDave
Joined: 03 Apr 2013
Last visit: 26 Jan 2020
Posts: 218
Own Kudos:
Given Kudos: 872
Location: India
Concentration: Marketing, Finance
GMAT 1: 740 Q50 V41
GPA: 3
GMAT 1: 740 Q50 V41
Posts: 218
Kudos: 276
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Ronak275
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

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.

Similar questions to practice?
User avatar
Romannepal
Joined: 12 Nov 2016
Last visit: 25 Sep 2017
Posts: 23
Own Kudos:
Given Kudos: 27
Location: Nepal
Concentration: Accounting, Economics
GPA: 3.63
WE:Account Management (Accounting)
Posts: 23
Kudos: 97
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
Ronak275
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

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.

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
Attachments

20170611_094035-1.jpg
20170611_094035-1.jpg [ 2.89 MiB | Viewed 5052 times ]

User avatar
pushpitkc
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,826
Own Kudos:
5,837
 [1]
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,826
Kudos: 5,837
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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!
User avatar
Romannepal
Joined: 12 Nov 2016
Last visit: 25 Sep 2017
Posts: 23
Own Kudos:
Given Kudos: 27
Location: Nepal
Concentration: Accounting, Economics
GPA: 3.63
WE:Account Management (Accounting)
Posts: 23
Kudos: 97
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
pushpitkc
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,826
Own Kudos:
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,826
Kudos: 5,837
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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,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!
User avatar
Regor60
Joined: 21 Nov 2021
Last visit: 15 Jun 2025
Posts: 498
Own Kudos:
Given Kudos: 435
Posts: 498
Kudos: 340
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ronak275
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

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
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,180
Own Kudos:
Posts: 37,180
Kudos: 1,000
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
102025 posts
PS Forum Moderator
638 posts