ME test series closed

Question

A computer has 32­bit instruction and 9­bit address. If there are 400 two address instructions then how many one address instructions can be formulated?

a. 2^ 14

b. 2 ^32 – 200

c. 2 ^14 – 400

d. (2^14 – 400) × 2^9 (Answer)

can anybody explain me how this option is correct?

Comments

400 opcodes have been used in two address instructions. So in order to formulate one address instructions, I will have to subtract those opcodes which were being attached with two address instructions. with 14 bits of opcodes, I can form 2^14 different instructions, after subtraction 400 from 2^14, with 9-bit address, the number of 1 address instructions will be (2^14 - 400) * 2 ^9