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How to determine if a simple graph of $7$ vertices where the degree of each vertex is at least $3$ is connected?

According to this answer we need to check that $|E|>{7-1\choose 2}=15$. If the degree of each vertex is at least $3$ then $\sum_{v\in V}deg(v)=21$. But $\sum_{v\in V}deg(v)$ cannot be odd so there must be at least one vertex with degree $4$ therefore $\sum_{v\in V}deg(v)=22$ at least. In such case $|E|=11$.

We see that $|E|<{7-1\choose 2}$ therefore the graph cannot be connected.

But on the other hand because the degree of one of the vertices is $4$ then there's a subgraph which has $5$ vertices. Suppose the other $2$ vertices are not connected but this cannot be because the degree of each vertex is at least $3$.

What am I doing wrong?

Yos
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    Note that the first part of the answer you link to gives only a sufficient condition for being connected, not a necessary one (despite the question there asking for both). – hmakholm left over Monica May 18 '18 at 10:41

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It seems easier to me just to say: If every vertex has at least 3 neighbors, then a connected component will have at least 4 vertices in it. 7 vertices is therefore not enough to have two connected components.

hmakholm left over Monica
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