1. **Problem Statement:** Determine which of the given sequences are circuits in the graph with vertices $V_1, V_2, V_3, V_4, V_5$ and edges $a(V_1\text{-}V_2), b(V_2\text{-}V_3), c(V_4\text{-}V_5), d(V_1\text{-}V_4), e(V_2\text{-}V_5)$. A circuit is a closed path where the start and end vertices are the same, and each consecutive pair of vertices is connected by an edge. 2. **Check each sequence:** - Sequence 1: $v_3 v_2 v_1 v_4 v_5 v_3$ - Edges: $V_3\to V_2$ (edge $b$), $V_2\to V_1$ (edge $a$ reversed), $V_1\to V_4$ (edge $d$), $V_4\to V_5$ (edge $c$), $V_5\to V_3$ (no direct edge). This sequence is **not** a circuit because $V_5$ and $V_3$ are not directly connected. - Sequence 2: $v_4 v_5 v_2 v_3 v_2 v_1 v_4$ - Edges: $V_4\to V_5$ (edge $c$), $V_5\to V_2$ (edge $e$ reversed), $V_2\to V_3$ (edge $b$), $V_3\to V_2$ (edge $b$ reversed), $V_2\to V_1$ (edge $a$ reversed), $V_1\to V_4$ (edge $d$). - Starts and ends at $V_4$, all edges exist, so this is a **circuit**. - Sequence 3: $v_1 v_2 v_5 v_4 v_1$ - Edges: $V_1\to V_2$ (edge $a$), $V_2\to V_5$ (edge $e$), $V_5\to V_4$ (edge $c$ reversed), $V_4\to V_1$ (edge $d$ reversed). - Starts and ends at $V_1$, all edges exist, so this is a **circuit**. - Sequence 4: $v_5 v_4 v_1 v_2 v_3$ - Edges: $V_5\to V_4$ (edge $c$ reversed), $V_4\to V_1$ (edge $d$ reversed), $V_1\to V_2$ (edge $a$), $V_2\to V_3$ (edge $b$). - Does not end at the starting vertex $V_5$, so this is **not** a circuit. 3. **Final answer:** The sequences that are circuits are: - $v_4 v_5 v_2 v_3 v_2 v_1 v_4$ - $v_1 v_2 v_5 v_4 v_1$