Two or three address memory, do I think correctly?

Question

Computers

4 bits opcode
24 bits instruction length
So using a two address system, the number of cells would be 2^((24-4)/2)=2^10=1024 right?
And using a three-address system 2^(20/3)=102 ?
And in that case, the maximum address that can be reached is the number of cells -1? that is, in the case of a two-address system 1024-1 =1023, and in the case of a three-address system: 102-1=101?