Realising a 3-variable function on a 4:1 multiplexer

Step-by-step derivation

1. Write out the minterms. With $A$ as MSB and $C$ as LSB, the required minterms are $m_1 = \overline{A}\,\overline{B}\,C$, $m_2 = \overline{A}B\overline{C}$, $m_4 = A\overline{B}\,\overline{C}$ and $m_7 = ABC$.
2. The select lines $AB$ choose which data input reaches the output, so group the eight possible input combinations into the four select cases and see what $F$ must be in each, as a function of $C$ alone.
3. $AB = 00$ (drives $I_0$): the two rows are $m_0$ ($C = 0$, $F = 0$) and $m_1$ ($C = 1$, $F = 1$). $F$ follows $C$, so $I_0 = C$.
4. $AB = 01$ (drives $I_1$): rows $m_2$ ($C = 0$, $F = 1$) and $m_3$ ($C = 1$, $F = 0$). $F$ is the complement of $C$, so $I_1 = \overline{C}$.
5. $AB = 10$ (drives $I_2$): rows $m_4$ ($C = 0$, $F = 1$) and $m_5$ ($C = 1$, $F = 0$). Again $I_2 = \overline{C}$.
6. $AB = 11$ (drives $I_3$): rows $m_6$ ($C = 0$, $F = 0$) and $m_7$ ($C = 1$, $F = 1$). $F$ follows $C$, so $I_3 = C$.
7. Collecting: $I_0 = C,\; I_1 = \overline{C},\; I_2 = \overline{C},\; I_3 = C$, which is option (A).
8. The insight worth keeping: $\sum m(1,2,4,7)$ is exactly the set of inputs with an odd number of 1s, so $F = A \oplus B \oplus C$. Recognising the parity function lets you write the four data inputs down immediately, because $A \oplus B \oplus C = (A \oplus B) \oplus C$ is $C$ when $A \oplus B = 0$ and $\overline{C}$ when $A \oplus B = 1$.