Question 342886
That would be the case if each of the letters was unique. 
You are overcounting.
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Think about a simpler setup.
Look at 3 distinct letters.
ABC
ACB
BAC
BCA
CAB
CBA
3!/(3-3)!=6 ways

Now look at only 2 distinct letters but 3 total letters.
AAC
ACA
CAA
Only 3 different patterns available.

If they were all the same then only AAA would be unique, 1 pattern. 
Think about it and repost if you need more help.