4.5 soccer teams are conducting a round-robin tournament (each pair of teams plays one match). It is known that Team A has played 3 matches, Team B has played more matches than Team A, Team C has played fewer matches than Team A, and Team D and Team E have played the same number of matches, but Team D and Team E have not played against each other. Therefore, the total number of matches played is $\qquad$
4.6.

Team B has played 4 matches. If Team C has only played 1 match, then Team C hasn't played against Team A, and Team A must have played against Teams D and E, so Teams D and E have each played 2 matches. Therefore, the total number of matches is
$$
(3+4+1+2+2) \div 2=6 \text {. }
$$

If Team C has played 2 matches, and Teams D and E have each played $x$ matches, then $3+4+2+x+x=9+2x$ is not an even number, leading to a contradiction.