What are all the possible arrangements of the word "EXAMINATION"? (probability)

do you use permutations here? or combinations? how is it done? just using those letters, obviously, most won't be words

the word has 11 letters of which 2 are A's, 2 are I's & 2 are N's

the simple formula for permutations of n objects when r1,r2, ..... are identical is

n! /(r1!*r2!*....) , so

# of arrangements = 11!/(2!*2!*2!) = 4,989,600 <-------