The exact question is:
A certain bridge club has a special rule to the effect that four members may play together only if no two of them have previously partnered each other. At one meeting fourteen members, each of whom has previously partnered five others, turn up. Three games are played and then proceedings come to a halt because of the club rule. Just as the members are preparing to leave, a new member, unknown to any one of them, arrives. Show that at least one more game can now be played.
If we construct a graph on $14$ vertices representing the $14$ people, connecting $v_i$ and $v_j$ if and only if person $i$ and $j$ have not been partnered with each other, we have the following equivalent formulation:
Show that if $|G| = 14$ and $G$ is a $8$-regular graph, if after removing $3$-copies of $K_4$ from it, $G$ does not contain another $K_4$, then this new graph contains a triangle.
It suffices to find a triangle in this new graph because it implies that the $3$ people involved in the triangle has not been paired up with one another. These $3$ people, along with the additional member, can now play a new game.
So far, these are the results I've established.
Before removing the $3$ copies of $K_4$, $e(G) = 56$.
After removing the $3$ copies of $K_4$, $e(G) = 56 - 3 \times 6 = 38$.
If $G$ does not contain a triangle, then by Mantel's theorem, $e(G) = 14^2/ 4 = 49$. This doesn't give a contradiction as hoped, since $49 > 38$.
I've also tried using the stronger version of Mantel's theorem, which is $e(G) \leq \alpha(G) \beta(G)$, where $\alpha(G)$ is the cardinality of the maximum independent set of $G$, and $\beta(G)$ is the cardinality of the minimum vertex cover of $G$. But I can't find a way to link it with the fact that $G$ minus three copies of $K_4$ no longer contains another $K_4$.
I've also established that after the removal of three copies of $K_4$, there still will be at least two vertices of degree $8$, and the minimum degree of the new graph is at leasat $2$.
Any hints are appreciated!