Binary multiplication — compute three 4-bit × 4-bit products Multiply these binary pairs independently and write each 8-bit product in order: (1010 × 0011), (1011 × 0111), (1001 × 1010).

Introduction / Context:Binary multiplication underpins arithmetic units in processors and programmable logic. Here you multiply three 4-bit operands by three 4-bit operands and present each 8-bit product, testing understanding of place value and partial products in base 2.

Given Data / Assumptions:

• P1: 1010 × 0011
• P2: 1011 × 0111
• P3: 1001 × 1010
• Unsigned binary; output each product as 8 bits with a space every 4 bits for readability.

Concept / Approach:

Convert to decimal to verify, or perform shift-and-add: multiplying by 2^k is a left shift by k. Sum partial products and limit to 8 bits for 4×4. Finally, sanity-check by converting back to decimal.

Step-by-Step Solution:

Verification / Alternative check:

Shift-add for P1: 1010 × (0011) = 1010 + (1010 << 1) = 1010 + 10100 = 11110 → zero-extend to 0001 1110. Similar checks confirm P2 = 0x4D, P3 = 0x5A.

Why Other Options Are Wrong:

• Option a: last group 0101 1011 = 91, not 90.
• Option b: second group 0100 1100 = 76, not 77.
• Option d: first group 0001 1101 = 29, not 30.
• Option e: multiple groups deviate from correct decimal equivalents.

Common Pitfalls:

• Forgetting that binary multiplication by 2 is a left shift.
• Dropping carries when summing partial products.

Final Answer:

0001 1110 0100 1101 0101 1010