Math Is Fun Forum

windwarner

Member

Registered: 2009-02-25

Posts: 1

Combinations

I was trying to figure out how many permuations there are of 13 things taken
three at a time where order does not matter and I believe it is 13! divided by
(3! x 10!)?  Right?  Now what if we add an additional constraint: repetitions of
the same two objects/numbers are not allowed. i.e., once you have 123, you can't
have 124, 125, 126, etc. or 234, 235, etc.
> Example - you have 13 people and you are going to have them all meet each other in groups of
three but you don't want any two people to see each other in a group more than
once.  What would the formula for that be?  Seems like that might be a fun
problem.

LampShade

Member

Registered: 2009-02-22

Posts: 23

Re: Combinations

You said permutations when you meant combinations.

The answer to your question is as follows:

The answer is shown below.

If order does not matter (i.e., 12 is the same as 21) then you are looking at a combination, and the formula is

Last edited by LampShade (2009-02-25 14:25:49)

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--  Boozer