Combinatorics??

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Combinatorics??

by quitaskinme » Sat Oct 31, 2015 10:52 pm
There are 10 children in a company's day-cre center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?

A)100
B)90
C)50
D)45
E)25
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by theCEO » Sun Nov 01, 2015 2:37 am
quitaskinme wrote:There are 10 children in a company's day-cre center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?

A)100
B)90
C)50
D)45
E)25
Number of ways to choose the 1st person: 10
Number of ways to choose the 2nd person: 9
Total combinations = 10 x 9 = 90 if order matters

Since question asks for different pairs we have to remove the duplicates which is 1/2 of 90
90 - (0.5x90) = 45 or
90(1-1/2) = 90(0.5) = 90/2 = 45

Ans = d

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by Brent@GMATPrepNow » Sun Nov 01, 2015 7:20 am
quitaskinme wrote:There are 10 children in a company's day-cre center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?

A)100
B)90
C)50
D)45
E)25
Another approach:

Let's say that every child is paired with another child and those to children play the game.

After all of the games have been played, we ask each of the 10 children, "How many different people did you play with?"
We'll find that EACH CHILD played with 9 other children, which gives us a total of 90 games (since 10 x 9 = 90).

From here we need to recognize that every pair of children has been counted TWICE. For example, if Child A and Child B play a game together, then Child A counts it as a game, AND Child B also counts it as a game. Of course only one game occurred.

To account for the duplication, we'll divide 90 by 2 to get 45

Answer: D

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by Brent@GMATPrepNow » Sun Nov 01, 2015 12:44 pm
Here's a related (and harder) question - https://www.beatthegmat.com/how-many-han ... 78283.html

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