**Question:**

For a 16 bit word with 6 bits for an opcode

- How many different instructions could I fit into the instruction set?
- What is the largest number that I could use as data?

**Answer:**

- Number of instructions: 2
^{6}= 64 - largest operand: 2
^{10}– 1 = 1023

Source

My question is how did they calculated 2^{10} from 16 bit word?

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If a number takes up 16 bits, and you use 6 of them for something, that leaves 10 bits.

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