Posted by **Anna** on Monday, April 4, 2011 at 5:29pm.

Please help-I'm really mixed up

How do I find the permutations of the first 12 letters taken 3 at a time?

I have a ton of these to do and really don't understand them at all-

- Algebra II-Please help!! -
**Damon**, Monday, April 4, 2011 at 5:38pm
permutations:

That means all the arrangements of three made from the 12 when order is important. For example

ABC and ACB and BAC and BCA and CAB and CBA are all different.

The formula for that is

n!/[n-r]!

here n = 12 and r = 3

so we have

12!/9! which is 12*11*10 = 132

That is what you asked.

To answer what is coming next, combinations of n taken r at a time where order is NOT important are called indeed combinations and the formula is logically enough the permutation formula divided by r!

C(n,r) = n!/[ r! (n-r)! ]

- Algebra II-Please help!! -
**Anna**, Monday, April 4, 2011 at 5:54pm
Thank you for the explanation-it actually makes sense now

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