, 2011

Imagine the fencers as nodes in a graph. There is an edge between two nodes if those fencers have duelled. If everyone is to have duelled a different number of opponents, then since there are only 10 fencers, they must have each duelled a distinct number from $\{0, 1, 2, \ldots, 8, 9\}$. In any graph, the sum of all the degrees of each node is even since each edge is counted twice. However, the sum of all the degrees for the 10 fencers would have to be $1 + 2 + \dots + 8 + 9 = \frac{9 \times 10}{2} = 45$ which is odd. Hence such a state cannot be achieved. Furthermore, note that an easier solution is to just note that you can't simultaneously use $0$ and $9$.