M06-11

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:

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