Algebra IIPlease help!!
posted by Anna on .
Please helpI'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

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!/[nr]!
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! (nr)! ] 
Thank you for the explanationit actually makes sense now