Answer from cs61c-ew (Joo-Rak Son 16103505) for Question 3
Total of six wires would be necessary for this module. First, looking at AB!C, since the logic gates only has two inputs, must be a combination of three logic gates: (1) AND A and B, (2) NOT C, (3) AND results from 1 and 2. So this requires two intermediary wires. The same applies to AC!B. Lastly, to combine these two terms together by taking the results from both sets of logic gates, we need two more wires. In total, we need six.