(a) graph $k_4$ (complete graph on 4 vertices)
☐ bipartite
☐ regular
☐ complete
☐ has a cycle
☐ complete bipartite

(b) graph $c_6$ (cycle graph on 6 vertices)
☐ has a cycle
☐ complete bipartite
☐ bipartite
☐ complete
☐ regular

(c) graph $p_5$ (path graph on 5 vertices)
☐ complete bipartite
☐ complete
☐ regular
☐ has a cycle
☐ bipartite

Question

(a) graph $k_4$ (complete graph on 4 vertices)
  ☐ bipartite
  ☐ regular
  ☐ complete
  ☐ has a cycle
  ☐ complete bipartite

(b) graph $c_6$ (cycle graph on 6 vertices)
  ☐ has a cycle
  ☐ complete bipartite
  ☐ bipartite
  ☐ complete
  ☐ regular

(c) graph $p_5$ (path graph on 5 vertices)
  ☐ complete bipartite
  ☐ complete
  ☐ regular
  ☐ has a cycle
  ☐ bipartite

Accepted Answer

# Explanation:
## Part (a) - $ K_4 $ (Complete graph on 4 vertices)
### Step 1: Recall Definitions
- **Bipartite**: A graph whose vertices can be divided into two disjoint sets such that no two graph vertices within the same set are adjacent. A complete graph with $ n \geq 3 $ vertices is not bipartite (since it contains odd-length cycles, e.g., $ K_3 $ is a triangle, a 3 - cycle). For $ K_4 $, it has cycles of length 3, 4, etc. So $ K_4 $ is not bipartite.
- **Regular**: A graph where each vertex has the same degree. In $ K_4 $, each vertex is connected to $ 4 - 1=3 $ other vertices, so degree of each vertex is 3. Thus, $ K_4 $ is 3 - regular.
- **Complete**: A complete graph $ K_n $ is a graph where every pair of distinct vertices is connected by a unique edge. By definition, $ K_4 $ is complete.
- **Has a cycle**: A complete graph with $ n\geq 3 $ has cycles. For $ K_4 $, we can find cycles like $ v_1 - v_2 - v_3 - v_1 $ (a 3 - cycle) or $ v_1 - v_2 - v_3 - v_4 - v_1 $ (a 4 - cycle). So it has a cycle.
- **Complete bipartite**: A complete bipartite graph $ K_{m,n} $ has partitions of size $ m $ and $ n $, and every vertex in one partition is connected to every vertex in the other. $ K_4 $ is not bipartite (as established earlier), so it is not complete bipartite.

So for $ K_4 $, the correct options are: Regular, Complete, Has a cycle.

## Part (b) - $ C_6 $ (Cycle graph on 6 vertices)
### Step 1: Recall Definitions
- **Has a cycle**: By definition, a cycle graph $ C_n $ is a graph that consists of a single cycle of length $ n $. So $ C_6 $ has a cycle.
- **Complete bipartite**: A cycle graph with even length $ n = 2k $ is bipartite (and in fact, for $ n = 6 $, $ C_6 $ is $ K_{3,3} $? No, $ C_6 $ can be partitioned into two sets $ A=\{v_1, v_3, v_5\} $ and $ B = \{v_2, v_4, v_6\} $, and it is a bipartite graph. A complete bipartite graph $ K_{m,n} $ has all possible edges between the two partitions. $ C_6 $ is bipartite but not complete bipartite (since in $ K_{3,3} $, each vertex in $ A $ is connected to each vertex in $ B $, but in $ C_6 $, $ v_1 $ is connected to $ v_2 $ and $ v_6 $, not to $ v_4 $ or $ v_5 $ etc.). Wait, actually $ C_6 $ is bipartite and is a cycle graph with even length. Also, in $ C_6 $, each vertex has degree 2.
- **Bipartite**: A cycle graph with even length is bipartite. Since $ 6 $ is even, $ C_6 $ is bipartite.
- **Complete**: A complete graph has all possible edges. In $ C_6 $, not all pairs of vertices are connected (e.g., $ v_1 $ and $ v_3 $ are not adjacent), so it is not complete.
- **Regular**: A graph where each vertex has the same degree. In $ C_6 $, each vertex has degree 2, so it is 2 - regular.

So for $ C_6 $, the correct options are: Has a cycle, Bipartite, Regular.

## Part (c) - $ P_5 $ (Path graph on 5 vertices)
### Step 1: Recall Definitions
- **Complete bipartite**: A path graph is a tree (connected acyclic graph). A complete bipartite graph has many edges, while a path graph has a linear structure. So $ P_5 $ is not complete bipartite.
- **Complete**: A complete graph has all possible edges. In $ P_5 $, vertices are connected in a line, so not all pairs are adjacent (e.g., the first and third vertices are not adjacent), so it is not complete.
- **Regular**: In a path graph $ P_n $, the two end - vertices have degree 1, and the internal vertices (if $ n>2 $) have degree 2. So in $ P_5 $, the first and fifth vertices have degree 1, and the second, third, fourth have degree 2. So it is not regular (since degrees are not all the same).
- **Has a cycle**: A path graph is a tree, and trees are acyclic (no cycles). So $ P_5 $ does not have a cycle.
- **Bipartite**: A path graph is a tree, and all trees are bipartite. We can partition the vertices of $ P_5 $ (vertices $ v_1, v_2, v_3, v_4, v_5 $) into two sets: $ A=\{v_1, v_3, v_5\} $ and $ B=\{v_2, v_4\} $. No two vertices in $ A $ are adjacent, and no two vertices in $ B $ are adjacent. So $ P_5 $ is bipartite.

So for $ P_5 $, the correct option is: Bipartite.

# Answer:
(a) Regular, Complete, Has a cycle
(b) Has a cycle, Bipartite, Regular
(c) Bipartite

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QUESTION IMAGE

Question

Explanation:

Part (a) - \( K_4 \) (Complete graph on 4 vertices)

Step 1: Recall Definitions

Part (b) - \( C_6 \) (Cycle graph on 6 vertices)

Step 1: Recall Definitions

Part (c) - \( P_5 \) (Path graph on 5 vertices)

Step 1: Recall Definitions

Answer: