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Assume that there are 251 different opcode and 32 registers in the machine.Every instruction has 3 register as input and 1 register as output[opcode , R1 , R2 , R3 , R4].The number of bits to encode an instruction

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Number of different opcodes given  =  251

Hence number of bits needed to represent opcode  =   ceil ( log2251 )    =   8

Number of registers given    =   32

Hence number of bits needed to represent register  =   log232    =    5

Hence number of bits needed for instruction   =   8 + 4 * 5 [ Bcoz one opcode field and 4 register fields ]

                                                                  =   28 bits

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