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Ronak275
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Ben and Shari have invited three other couples to their apar 03 Jul 2014, 12:51
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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
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A
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Ben and Shari have invited three other couples to their apar 03 Jul 2014, 13:43
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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
Show more

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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Ben and Shari have invited three other couples to their apar 04 Oct 2016, 22:20
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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
Show more

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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Ben and Shari have invited three other couples to their apar 06 Jul 2014, 05:43
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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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Ben and Shari have invited three other couples to their apar 06 Jul 2014, 10:26
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antra.nith@gmail.com
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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Ben and Shari have invited three other couples to their apar 07 Jul 2014, 11:12
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@Bunuel:

Thanks a lot!
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Ben and Shari have invited three other couples to their apar 04 Oct 2016, 20:03
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hi could you explain this in detail please
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Ben and Shari have invited three other couples to their apar 04 Oct 2016, 20:35
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Bunuel
antra.nith@gmail.com
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.
Show more

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
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Ben and Shari have invited three other couples to their apar 04 May 2017, 07:49
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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
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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
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Ben and Shari have invited three other couples to their apar 04 May 2017, 10:32
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antra.nith@gmail.com
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
Show more

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?
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Ben and Shari have invited three other couples to their apar 08 May 2017, 07:13
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KM2018AA
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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
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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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Ben and Shari have invited three other couples to their apar 28 May 2017, 02:39
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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
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Ben and Shari have invited three other couples to their apar 10 Jun 2017, 07:32
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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.
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Ben and Shari have invited three other couples to their apar 10 Jun 2017, 21:04
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Bunuel
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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.
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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.

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Ben and Shari have invited three other couples to their apar 11 Jun 2017, 00:59
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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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Ben and Shari have invited three other couples to their apar Updated on: 11 Jun 2017, 06:23
Originally posted by Romannepal on 11 Jun 2017, 04:04.
Last edited by Romannepal on 11 Jun 2017, 06:23, edited 1 time in total.
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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!
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Thanks. However, you have just interchanged the options. Doesn't that mean the same team pair anyway? Please clarify.
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Ben and Shari have invited three other couples to their apar 11 Jun 2017, 10:15
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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.
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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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Ben and Shari have invited three other couples to their apar 29 Jun 2022, 14:16
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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
Show more

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

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Ben and Shari have invited three other couples to their apar 04 Aug 2024, 07:46
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Hello from the GMAT Club BumpBot!

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