1. Problem: Give an example of a simple graph with 12 vertices and 35 edges. Step 1: Recall that a simple graph with $n$ vertices can have at most $\frac{n(n-1)}{2}$ edges. Step 2: For $n=12$, maximum edges = $\frac{12 \times 11}{2} = 66$. Step 3: Since 35 is less than 66, such a graph exists. Step 4: One example is a graph where 11 vertices form a complete graph $K_{11}$ with $\frac{11 \times 10}{2} = 55$ edges, but we only need 35 edges, so we can take a subgraph of $K_{12}$ with 35 edges by selecting edges accordingly. Note: Drawing is not possible here. 2. Problem: Does there exist a bipartite graph with vertex degrees 4, 3, 3, 2, 2, 2? Step 1: In a bipartite graph, sum of degrees in one part equals sum in the other. Step 2: Sum of degrees = $4 + 3 + 3 + 2 + 2 + 2 = 16$. Step 3: For bipartite graph, vertices split into two parts $U$ and $V$ such that sum of degrees in $U$ equals sum in $V$. Step 4: Try to partition degrees into two sets with equal sum 8. Step 5: Possible partition: $\{4, 2, 2\}$ and $\{3, 3, 2\}$ both sum to 8. Step 6: Hence, such a bipartite graph exists. 3. (a) Can a complete bipartite graph be a complete graph? Step 1: A complete bipartite graph $K_{m,n}$ has edges between two disjoint sets of vertices. Step 2: A complete graph $K_p$ has edges between every pair of vertices. Step 3: $K_{m,n}$ is complete graph only if one part has 1 vertex, i.e., $K_{1,n}$ which is a star graph. Step 4: So, only $K_{1,1} = K_2$ is both complete bipartite and complete graph. 3. (b) Can a complete graph be a complete bipartite graph? Step 1: Complete graph $K_p$ has edges between all pairs. Step 2: Complete bipartite graph $K_{m,n}$ has no edges within parts. Step 3: So, $K_p$ can be complete bipartite only if $p=2$ (i.e., $K_2 = K_{1,1}$). Step 4: For $p>2$, $K_p$ is not bipartite. 4. Find number of vertices: (a) 16 edges and vertices of degree 2. Step 1: Sum of degrees = $2 \times n$. Step 2: Sum of degrees = $2 \times$ number of edges = $2 \times 16 = 32$. Step 3: So, $2n = 32 \Rightarrow n=16$ vertices. (b) 21 edges, 3 vertices of degree 4, rest degree 3. Step 1: Let total vertices be $n$. Step 2: Sum of degrees = $2 \times 21 = 42$. Step 3: Sum of degrees = $3 \times 4 + (n-3) \times 3 = 12 + 3n - 9 = 3n + 3$. Step 4: Set equal: $3n + 3 = 42 \Rightarrow 3n = 39 \Rightarrow n=13$ vertices. (c) 24 edges and all vertices same degree $d$. Step 1: Sum of degrees = $2 \times 24 = 48$. Step 2: Let number of vertices be $n$, all degree $d$. Step 3: Sum of degrees = $n \times d = 48$. Step 4: Possible pairs $(n,d)$ satisfy $n d = 48$ with $d \leq n-1$. Step 5: Examples: $(n,d) = (12,4), (8,6), (6,8)$ but $d \leq n-1$ so $(6,8)$ invalid. Step 6: So possible $n=12$ with $d=4$ or $n=8$ with $d=6$. Final answers: 1. Exists, example subgraph of $K_{12}$ with 35 edges. 2. Yes, bipartite graph exists. 3. (a) Only $K_{1,1}$ is both. 3. (b) Only $K_2$ is both. 4. (a) 16 vertices. 4. (b) 13 vertices. 4. (c) $n=12$ with $d=4$ or $n=8$ with $d=6$.