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10 children, each group has 2 children. how many groups pos?

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by phoenixhazard » Sun Oct 17, 2010 1:09 pm
Sounds like such an easy question but I can't find how to do it fast and simple.

"There are 10 children in a day care, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?"
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by shovan85 » Sun Oct 17, 2010 1:26 pm
phoenixhazard wrote:Sounds like such an easy question but I can't find how to do it fast and simple.

"There are 10 children in a day care, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?"
This time I am correct dude (think so) ;)
It is a Combinatorics problem.
Out of 10 Children 2 can be selected in C(10,2) ways.

C(10,2) = 10!/(2!*8!) = 45
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by phoenixhazard » Sun Oct 17, 2010 9:03 pm
awesome thanks

so its (Total Number of students)/(Number per group)*(other options)?
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by shovan85 » Mon Oct 18, 2010 2:25 am
phoenixhazard wrote:awesome thanks

so its (Total Number of students)/(Number per group)*(other options)?
This is a formula in General:

Let n be the total number of choices
r be the choices to be selected from n

If the arrangement of r choices does not matter then use Combinatorics:
C(n,r) = n!/[r!*(n-r)!]

If the arrangement of r choices matters then use Permutations:
P(n,r) = n!/(n-r)!

One Q for you ;)

Q: "There are 10 children in a day care, and 8 children are to be selected to play a game. At most, how many different groups are possible?"
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by Geva@EconomistGMAT » Mon Oct 18, 2010 2:39 am
Nice one, Shovan. :)
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