A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. Find that number. The number of different ways that the letters of "possession" can be arranged is nothing.

A classic counting problem is to determine the number of different ways that the letters of "possession" can be arranged. We have, word "POSSESSION" which have 10 letters! as P,O,S,S,E,S,S,I,O,N . We have to find the number of ways letters of word "possession" can be arranged , in order to do that we will use simple logic as:

At first, we have a count of 10 letters and 10 places vacant to fill letters.

For first place we have a choice of 10 letters , after putting some letter from all 10 letters in first place now we are left with 9 places and 9 letters having 9 choices , similarly we'll be having 8 places , 8 letters and 8 choices and so on...... Therefore, Number of possible ways to arrange letters of word "possession" = [tex]10.9.8.7.6.5.4.3.2.1[/tex] = [tex]10![/tex] ( 10 factorial ) , but there are 4 letters "s" are repeated and 2 letters "o" are repeated so we need to eliminate the similar combinations ∴ Number of possible ways = [tex]\frac{10!}{4!(2!)}[/tex] = 75600.

∴Number of possible ways to arrange letters of word "possession" = 75600