Aclassic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. find that number?

You are in the wrong subject, but I'll answer the question, but it really belongs in math. The answer is

10!/(4! 2!)

Explanation

If the letters were all different and you were to use all of them, then the number of arrangements for 10 letters is 10! (read that as 10 factorial. It means    10*9*8*7*6*5*4*3*2*1

There are 4s. It does not matter which one you pick to go where so you get rid of the duplicates that reflects the fact that you could pick any one of the 4 s. So you divide by 4!

There are 2 o s . You divide by 2 for the same reason.

did you ever get an answer? i'm on the same question : (

we need to see the paragraphs in order to answer this.