Question 1131701
Breaking down the numbers into two cases, we can add the two counts:
1.  Those with one "I" and
2.  Those with two "I"s <br>

With one "I":
  {{{ (6C4)*4! = 15*24 = 360 }}}<br>

With two "I"s:
  {{{ (5C2)*(3C2)*2! + (5C2)*3! = 60 + 60 = 120 }}}<br>

(First term is where the two I's are separated by at least one letter and
the 2nd term is where the two I's are together, the 5C2 arises because we 
have already selected two Is and now are free to choose 2 of the remaining 
5 letters) <br>

Adding gives 360+120 = {{{ highlight( 480 ) }}} distinct permutations


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Caution: Tutor MathLover's answer does not take into account the two "I"s.
 Therefore the answer she has arrived at has overcounted the number of 
unique permutations.