How many different rearrangements of the letters in the word POSSESSION are possible if the letter E must come before the letter P?

Question

Permutation Question

Answer 1

the number of each letter: 1 E, 1 I, 1 N, 2 Os, 1 P, and 4 S.

Therefore, as possession has 10 letters, the number of possible arrangements is

10!/(1!1!1!2!1!4!) = 

then, half the time, e comes before p, and half the time the other, so the solution is 1/2 * 10!/(1!1!1!2!1!4!) = 37800