pesfunk wrote:A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420
B. 2520
C. 168
D. 90
E. 105
OA E
This is how I did it,
1st pair = 8 * 7 = 56
2nd pair = 6 * 5 = 30
3rd pair = 4 * 3 = 12
4th pair = 2 * 1 = 2
Total = 100 Which is not the answer. What did I do wrong here ?
Here's one approach:
8 people.
1st person can be paired with any one of the 7 remaining people = 7 possible pairs.
6 people left.
Next person can be paired with any one of the 5 remaining people = 5 possible pairs.
4 people left.
Next person can be paired with any one of the 3 remaining people = 3 possible pairs.
2 people left = 1 possible pair.
Thus, we have 7 choices for the first pair, 5 choices for the second pair, 3 choices for the third pair, and 1 choice for the final pair.
To combine the number of choices we have for each pair, we multiply: 7*5*3*1 = 105.
The correct answer is
E.
Please note the following distinction:
Given 4 people ABCD, the number of ways in which 2 different pairs can be
assigned to 2 different projects = 4C2*2C2 = 6.
Please note the significance of the word
assigned. The result above counts the following:
AB CD
CD AB
AC BD
BD AC
BC AD
AD BC
6 possible ways to
assign the pairs.
But notice that some of the assignments above involve duplicate pairings.
If all we want is to count the number of ways to
divide the 4 people, AB CD is no different from CD AB.
Thus, when counting the number of ways to
divide the 4 people into 2 pairs, because we don't care about the order of the pairings, we need to divide by the number of ways that the 2 pairs can be arranged (2!):
Number of ways to divide 4 people into 2 pairs = (4C2)*(2C2)/2! = 3.
The result above counts all the ways in which the 4 people can be divided into pairs:
AB CD
AC BD
BC AD
3 possible ways to
divide the 4 people into pairs.
Thus, in the problem above, the number of ways that the 8 friends can be divided into 4 pairs = (8C2*6C2*4C2*2C2)/4! = 105.
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