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There is complete graph $K_{n}$

Find formula for maksimal number of disjoint to each other $k$-cliques in $K_{n}$. Let denote it as $A(n,k)$

I did some research and i computed a formula: $B(n,k)=\lfloor \frac{{n\choose 2}}{{k\choose 2}}\rfloor$.

I thought that equality $A(n,k)=B(n,k)$ holds for every $n,k$, but i was wrong.

For example:

  • $A(4,3)=1=B(4,3)-1$
  • $A(5,3)=2=B(5,3)-1$
  • $A(6,3)=4=B(6,3)-1$
  • $A(6,4)=1=B(6,4)-1$

Is this true that $A(n,k)=B(n,k)-1$?

Regards

mkultra
  • 1,410
  • Does "disjoint to each other" mean edge-disjoint? Obviously there can't be more (or less) than $\lfloor{n/k}\rfloor$ if you require them to be vertex-disjoint. – mjqxxxx Apr 01 '19 at 13:38
  • I mean that cliques schould be edge-disjointed – mkultra Apr 01 '19 at 14:31

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