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There are 250 2-address instructions. There are only a total of 256 2-address instructions allowed if we have 32-bit instructions (two addresses take up 24 bits, leaving only 8 bits for the opcode). Looking at the 8 bit opcode, assume bit patterns 00000000 (0) through 11111001 (249) are used for the 250 two-address instructions. Then there are 6 bit patterns left for one address instructions. However, each one of these can use the remaining 12 bits gained from having only one operand, so we have 6 * 212. 00000000 11111001 11111010 11111010 ... 11111111 11111111 ______________________________________________________________________________ xxxxxxxxxxxx xxxxxxxxxxxx 000000000000 111111111111 000000000000 xxxxxxxxxxxx 111111111111 xxxxxxxxxxxx xxxxxxxxxxxx xxxxxxxxxxxx xxxxxxxxxxxx xxxxxxxxxxxx 250 2-address instructions 212 212 6 * 212 1-address instructions

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Q: A computer has 32 bit instructions and 12 bit addresses. If there are 250 two address instructions how many one address instructions can be formulated?
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