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nakib77, both are correct.
1) Consider each letter seperately,
- There is 1 A, we have 6 places to put this in - 6C1
- There are 2 B's, we have 5 places for this - 5C2
- There are 3 C's, we have 3 places - 3C3
Total possible = 6C1*5C2*3C3 = 6*10*1 = 60 ways
2) Consider all the letters together,
Total possible arrangement for 6 unique letters - 6!
Since all are not unique here - We have 3 C's and 2 B's,
It is 6!/2!3!