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03 Jul 2014, 12:51
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Question Stats:
24% (02:26) correct
76%
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based on 309
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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
(A) 11
(B) 14
(C) 22
(C) 35
(D) 70
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
(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.
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
(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
