Mission2012 wrote:How many different ways can a group of 6 people be divided into 3 teams of 2 people each?
(A) 4
(B) 9
(C) 15
(D) 24
(E) 36
Approach 1:
From the 6 people, the number of ways to choose 2 for the first team = 6C2 = (6*5)/(2*1) = 15.
From the 4 remaining people, the number of ways to choose 2 people for the second team = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining people, the number of ways to choose 2 people for the third team = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
15*6*1.
Since the ORDER of the teams doesn't matter -- AB-CD-EF is the same way of dividing the 6 people as CD-EF-AB -- the result above must be divided by the number of ways to ARRANGE the 3 teams (3!):
(15*6*1)/(3*2*1) = 15.
The correct answer is
C.
We could also GRIND IT OUT.
Let the 6 people be A, B, C, D, E and F.
Person A must be paired with one of the 5 other people.
Options:
AB, AC, AD, AE, AF.
Groupings that can be combined with the first red pair (
AB):
CD-EF
CE-DF
CF-DE.
Total options = 3.
The same reasoning can be applied to all 5 of the red pairs above.
Since there will be 3 options for each of the 5 red pairs, the number of ways to divide the 6 people into pairs = 3*5 = 15.
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