\section*{Task 1 - 091211}

At an evening event, each of the gentlemen present danced with exactly three ladies, and with each exactly once. When all participants were still sitting together in a cozy circle after the dance and reviewing the evening, it was noted that each of the ladies present had danced with exactly two gentlemen, and with each exactly once. Furthermore, it was observed that any two of the gentlemen had had exactly one common dance partner during the evening.

The number of all ladies and gentlemen present at this event is to be determined.
}

If the number of gentlemen present is denoted by \(h\) and that of the ladies by \(d\), then the number of dances performed on this evening is both \(3 h\) and \(2 d\), so that the relationship

\[
3 h=2 d
\]

holds between \(h\) and \(d\). A further relationship follows from this, that the number of pairs of gentlemen, namely \(\frac{1}{2}(h-1) h\), matches the number of ladies, so that

\[
\frac{1}{2}(h-1) h=d
\]

also holds. From (1) and (2), it follows that \(h=4, d=6\), so that only these numbers are in question for the solution.

That with these numbers there actually exists (and up to the arbitrary numbering of the participants, exactly one) solution that satisfies all three conditions is shown by the following list of 12 dance pairs, in which the gentlemen are numbered from 1 to 4 and the ladies from 1' to 6':

\[
11', 22', 33', 14', 25', 36', 21', 32', 13', 44', 45', 46'
\]

Here, as required, \(1, 2, 3\), and 4 each appear exactly three times and \(1', 2', \ldots, 6'\) each appear exactly twice.

\section*{Adapted from \([5]\)}