If the three E's are together, you might as well just count them as a single letter.
So, there are just 6! ways to arrange the letters.
Your 5!*6 is the same value. I assume you considered the 5! permutations of BRKLY times the 6 places you could stick the E's.
In how many arrangements of the letters BERKELEY are all three E's adjacent?
I think its (5!)*(6) i'm looking for some clarification.
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