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 it says memory is word addressable , and 256 word memory means memory size is 256 words.

also it is given 1 word = 20 bits , so, to address memory there require 8 bits. ( 28 = 256 )

20 bits instruction is divided into 8 +8 +4 where  8 bits for 1 address , 8 bits for 2 address and 4 bits for opcode.

With 20 bits for an instruction giving 220 possible encoding.

No. of encoding for 2 address instructions  =  8 * 28 * 2 = 2 19 

No. of encoding for 1 address instructions = 1984 * 28 

Then total number of possible 0-address instructions  =  220  -  { 2 19  +  (1984 * 28 ) } 

=  1048576   -  { 524,288 + (256 * 1984 ) }

= 1048576 -  ( 524288 + 507904 )

= 1, 048,576 - 1,032,192

=  16, 384 

The total number of '0' address instructions formulated is 16,384 .

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