ME TEST

Question

The format of double operand instruction of a CPU consist of $5$ bits opcode and $6$ bits for source and destination. $26$ double operand instructions and $184$ single operand instructions must be implemented. What will be the total number of zero operand instructions can be implemented?

How to solve such questions?

I always get confused.

Comments

6 bits for source and dest each or combined? I’m confused there :|

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I think each.

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Then

It will be $2^{17} - 26*2^{12} - 184*2^6 = 12800$ Is it correct?

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The answer given by @ashwin kulkarni is correct. When you have only zero operand instruction max can be 2^17. Now when 2 operand instruction takes some of it away then no of Kissimmee operand instructions taken away by 2 operand is 26*2^12(no of bits for register/memory) . Same for one operand comes to 184*2^6 . Subtracting with the total gives you no of zero operand instructions.