I need to prove that the directed graph that has 5 vertices where - three vertices have indegree 3, two have indegree 0; three vertices have outdegree 3, two have outdegree 0 - doesn't exist.
From my understanding, as the number of odd degrees vertices is odd - 3, and even the sum of degrees is 9, such graph shouldn't exist. But those zeroes kinda confuse me so I am not sure if sum & count prove it.
How to can I prove this?
Since there are $5$ vertices, and only two with in-degree $0$ or out-degree $0$, there must be at least one with both in-degree and out-degree equal to $3$. Let $x$ be such at vertex. Then there are $3$ vertices $y_1,y_2,y_3$ such that there is an edge from $x$ to $y_i, i=1,2,3.$
We have a contradiction already. Do you see it?