\section*{Problem 1 - 141211}

At the end of a large evening party, it turned out that none of the gentlemen present had danced with fewer than 10 and none with more than 12 ladies, while none of the ladies had danced with fewer than 12 and none with more than 14 gentlemen. No gentleman had danced with the same lady more than once. If every gentleman had danced with every lady exactly once, 480 dances would have taken place. Each dance performed by a gentleman with a lady counts as one dance. (If, for example, exactly 15 pairs dance simultaneously, this should be understood as 15 dances and not as 1 dance.)

a) Determine all possibilities for the number of ladies and gentlemen present that are consistent with these conditions.

b) Provide (in the simplest form of a rectangular scheme) one of the possible combinations of dance pairs that meets the conditions of the problem for the found numbers.
}

a) If the number of gentlemen is denoted by \(h\) and the number of ladies by \(d\), then \(h \cdot d = 480 \quad\) (1).

If the numbers of gentlemen who danced with 10, 11, and 12 ladies, respectively, are denoted by \(h_{1}, h_{2}, h_{3}\), and the numbers of ladies who danced with 12, 13, and 14 gentlemen, respectively, are denoted by \(d_{1}, d_{2}, d_{3}\), then

\[
h_{1} + h_{2} + h_{3} = h \quad ; \quad d_{1} + d_{2} + d_{3} = h
\]

while the number \(T\) of dances performed that evening can be given in two ways:

\[
T = 10 h_{1} + 11 h_{2} + 12 h_{3} \quad ; \quad T = 12 d_{1} + 13 d_{2} + 14 d_{3}
\]

From the transformations

\[
\begin{array}{lll}
T = 10 (h_{1} + h_{2} + h_{3}) + h_{2} + 2 h_{3} & ; & T = 12 (d_{1} + d_{2} + d_{3}) + d_{2} + 2 d_{3} \\
T = 12 (h_{1} + h_{2} + h_{3}) - 2 h_{1} - h_{2} & ; & T = 14 (d_{1} + d_{2} + d_{3}) - 2 d_{1} - d_{2}
\end{array}
\]

it follows that

\[
10 h < T \quad, \quad 14 d > T
\]

and thus \(12 d < T < 12 h\) and \(10 h < T < 14 d\), so \(d < h < \frac{7}{5} d\) summarized as

\[
d < h < \frac{7}{5} d
\]

Furthermore, due to the conditions \(h \geq 12, d \geq 10\). The factorizations of 480 that satisfy both conditions are

\[
10 \cdot 48 \quad, \quad 12 \cdot 40 \quad, \quad 15 \cdot 32 \quad, \quad 16 \cdot 30 \quad, \quad 20 \cdot 24
\]

From this, it follows from (1) that \(h = 24, d = 20\) must be the case, because only in this case is (2) also satisfied.

Thus, these values are the only possible numbers for the gentlemen and ladies.

b) With these numbers, all conditions of the problem are indeed satisfied, as the following example of an overview of dance pairings shows (see figure):

Gentleman No. \(\longrightarrow\)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1968.jpg?height=1251&width=1533&top_left_y=300&top_left_x=273)

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