K4 and Q3 are graphs with the following structures. Which one of the following…

Key idea: determine whether each graph can be drawn in the plane without edge crossings.

• K4: The complete graph on four vertices has v = 4 and e = 6. It admits a planar embedding (for example, draw a triangle for three vertices and place the fourth vertex inside, connecting it to the three outer vertices). Euler's formula holds (v − e + f = 2) with f = 4, so K4 is planar.

• Q3: The 3-cube (cube graph) has v = 8 and e = 12 and is the skeleton of a cube. It can be drawn in the plane without crossings (the usual planar drawing of a cube), and Euler's formula holds with f = 6. Therefore Q3 is planar.

• Conclusion: Both K4 and Q3 are planar graphs, so the correct statement is that both graphs are planar.