Answer:
Explanation:
[tex]\mathbf{F(A, B, C) = \bar A \bar B C + B \bar C + AB}[/tex]
[tex]\mathbf{F(A, B, C) = \bar A \bar B C + AB \bar C + \bar A B \bar C + ABC + AB \bar C}[/tex]
[tex]\mathbf{F(A, B, C) = 001,110,010,111,110}[/tex]
Hence;
[tex]\mathbf{F(A, B, C) = \sum m (1,2,6.7)}[/tex]
[tex]\mathbf{ A \ \ \ B \ \ \ C \ \ \ \ \ \ \ F } \\ \\ \mathbf{ 0 \ \ \ \ 0 \ \ \ \ 0 \ \ \ \ \ \ \ 0 } \\ \\ \mathbf{ 0 \ \ \ \ 0 \ \ \ \ 1 \ \ \ \ \ \ \ 1 }\to \ \ D_0 = C \\ \\ \mathbf{ 0 \ \ \ \ 1 \ \ \ \ 0 \ \ \ \ \ \ \ 1 } \to \ \ D_1 = \bar C \\ \\ \mathbf{ 0 \ \ \ \ 1 \ \ \ \ 1 \ \ \ \ \ \ \ 0 } \\ \\ \mathbf{ 1 \ \ \ \ 0 \ \ \ \ 0 \ \ \ \ \ \ \ 0 } \\ \\ \mathbf{ 1 \ \ \ \ 0 \ \ \ \ 1 \ \ \ \ \ \ \ 0 }[/tex]
[tex]\\ \\ \mathbf{ 1 \ \ \ \ 1 \ \ \ \ 0 \ \ \ \ \ \ \ 0 } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \to D_3 = 1 \ \ ; D_2 = 0 \\ \\ \mathbf{ 1 \ \ \ \ 1 \ \ \ \ 1 \ \ \ \ \ \ \ 1 }[/tex]
The 4-to-1 MUX and inverter is shown in the attached file below.