Handshake Parity Every person on earth has already shook hands with a certain number of people. Show that the number of people who shook other people's hands an odd number of times is even. What I thought: Denote people with vertices. Connect a pair of vertices (if they shook hand with each other) with an edge. Sum the number of edges each vertex is connected to (it's degree). This must be even as each edge is counted twice. It follows that there are even no of people with odd handshakes

Are you simply checking if your answer is totally correct? It is!

The argument you give for the sum of degrees is important in graph theory and you may have seen it expressed as

$$2E=\sum_1^V \rho_i$$

for a graph with $E$ edges and $V$ vertices.