it's been a while since i had permutations class, please correct me if i'm wrong
A C T I V A T E D
if 2 A's are separated by 5 letters then the word would be in the form: A (5 random letters) A (2 random letters)
If there were no repeated letters, the number of permutations would be 7! for 7 blank spaces
Since T is present twice, we divide it by 2!, which would mean number of permutations = 7!/2! = 2520
Similarly, the two A's can be separated by 6 and 7 letters too, total permutations = 2520 * 3 = 7560
forgive me if i am wrong i tried :)
https://www.quora.com/Given-the-word-STAPLE-how-many-different-arrangements-are-there-of-the-letters-such-that-A-and-E-are-separated-by-at-least-one-letter
this might help you
note in your case both letters are the same so divide by 2!
ACTIVATED has 9 letters
we need to arrange 2 A's so not counting A, there are 7 letters and 7! ways to arrange those.
However the letter T is repeteated twice, so we divide 7!/2!
Question says atleast 5, so we multiple 7!/2! with 3 for all arangements for 5 letter gap, 6 letter gap and 7 letter gap
thus answer is 7!/2! * 3
Discord