I'm not sure I am doing this right:
Using inclusion/exclusion principle
S = MATHISFUN has 9 letters so this have 9! permutations.
A = MATH , B = IS , C = FUN
$\#A = 6!$
$\#B = 8!$
$\#C = 7!$
$ \#(A \cap B) = 5! $
$ \#(A \cap C) = 4! $
$ \#(B \cap C) = 6! $
$ \#(A \cap B \cap C) = 3! $
So without this words we have
$ S \ (A \cup B \cup C) = 9! - (6! + 8! + 7!) - (5! + 4! +6!) + 3! $
Is it correct solution? I will be thankful for every help.
You are almost correct but it should be $9! - (6! + 8! + 7!) + (5! + 4! +6!) - 3!$ instead. If you look a t the venn diagram you will realize you need to alternate the sign for each inclusion/exclusion instead of doing it symmetrically over the four terms.