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16 Sep 2014, 00:27
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In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?
A. 90
B. 105
C. 168
D. 420
E. 2520
A. 90
B. 105
C. 168
D. 420
E. 2520
08 Oct 2014, 05:03
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Official Solution:
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?
A. 90
B. 105
C. 168
D. 420
E. 2520
A group of 8 people can be divided into 4 teams of 2 people each in \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=\frac{28*15*6*1}{24}=105\) different ways.
Above, \(C^2_8\) is the number of ways to choose 2 people for the first pair, \(C^2_6\) is the number of ways to choose 2 people for the second pair, \(C^2_4\) is the number of ways to choose 2 people for the third pair, and \(C^2_2\) is the number of ways to choose 2 people for the fourth pair.
We divide by 4! (the factorial of the number of pairs) because the order of the pairs does not matter. For example, if the 8 people are labeled 1, 2, 3, 4, 5, 6, 7, and 8, the pair-selection (1,2); (3,4); (5,6); (7,8) would be considered the same as (7,8); (1,2); (3,4); (5,6), as there is no specific order assigned to the pairs. And since \(C^2_8*C^2_6*C^2_4*C^2_2\) will give all 4! arrangements of (1,2); (3,4); (5,6); (7,8) pair-selection, we need to divide this and each other possible pair-selection by 4! to eliminate the essentially identical pair-selections.
An alternative way to think about this is as follows:
• For the first person, there are 7 possible choices for a teammate;
• For the next unpaired person, there are 5 possible choices for a teammate (as two people are already chosen);
• For the third unpaired person, there are 3 possible choices for a teammate (as four people are already chosen);
• For the fourth unpaired person, there is only one choice left for a teammate.
Multiplying these choices together gives us 7*5*3*1 = 105 different ways to divide the group of 8 people into 4 teams of 2 people each.
Answer: B
In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?
A. 90
B. 105
C. 168
D. 420
E. 2520
A group of 8 people can be divided into 4 teams of 2 people each in \(\frac{C^2_8*C^2_6*C^2_4*C^2_2}{4!}=\frac{28*15*6*1}{24}=105\) different ways.
Above, \(C^2_8\) is the number of ways to choose 2 people for the first pair, \(C^2_6\) is the number of ways to choose 2 people for the second pair, \(C^2_4\) is the number of ways to choose 2 people for the third pair, and \(C^2_2\) is the number of ways to choose 2 people for the fourth pair.
We divide by 4! (the factorial of the number of pairs) because the order of the pairs does not matter. For example, if the 8 people are labeled 1, 2, 3, 4, 5, 6, 7, and 8, the pair-selection (1,2); (3,4); (5,6); (7,8) would be considered the same as (7,8); (1,2); (3,4); (5,6), as there is no specific order assigned to the pairs. And since \(C^2_8*C^2_6*C^2_4*C^2_2\) will give all 4! arrangements of (1,2); (3,4); (5,6); (7,8) pair-selection, we need to divide this and each other possible pair-selection by 4! to eliminate the essentially identical pair-selections.
An alternative way to think about this is as follows:
• For the first person, there are 7 possible choices for a teammate;
• For the next unpaired person, there are 5 possible choices for a teammate (as two people are already chosen);
• For the third unpaired person, there are 3 possible choices for a teammate (as four people are already chosen);
• For the fourth unpaired person, there is only one choice left for a teammate.
Multiplying these choices together gives us 7*5*3*1 = 105 different ways to divide the group of 8 people into 4 teams of 2 people each.
Answer: B
28 Dec 2017, 01:23
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For similar questions check DISTRIBUTING ITEMS/PEOPLE/NUMBERS... (QUESTION COLLECTION)
Hope it helps.
Hope it helps.
