Artificial intelligent assistant

Prove a relation related to sets In a city, among each pair of people, there can be exactly one of k different relationships (relationships are symmetric). A crowd is a set of three people in which every pair have the same relation. Let $R_k$ denote the smallest number of people in the city such that the city always contains a crowd, then prove that $R_k$ $\leq$ [$ek!$] + 1 Thanks in advance. Please help me in putting under proper tag.

The "ramsey-theory" tag would be appropriate. (The standard formulation for your problem is in terms of monochromatic triangles in an edge-coloring of a complete graph.) The usual notation for the greatest integer or floor function nowadays is $\lfloor x\rfloor$ not $[x]$. Here are your hints:

For $k\in\Bbb N$, let $A_k=\lfloor k!e\rfloor=\frac{k!}{0!}+\frac{k!}{1!}+\dots+\frac{k!}{k!}$. Then $A_1=2$, and $A_k=kA_{k-1}+1$ for $k>1$. Use induction on $k$ to show that, if there are $k$ possible relationships, then any city with $A_k+1$ people must contain a crowd. (Consider one citizen, and use the Pigeonhole Principle to show that he is in the _same_ relationship to at least $A_{k-1}+1$ of his fellow citizens; then apply the Induction Hypothesis.) Conclude that $R_k\le A_k+1=\lfloor k!e\rfloor+1$.

By the way, the equality $R_k=\lfloor k!e\rfloor+1$ holds only for $k=1,2$ and $3$.

xcX3v84RxoQ-4GxG32940ukFUIEgYdPy fd6372cae0c714b2f0d553b61a9c4528