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I encountered a problem in Applied Combinatorics as detailed here which presents an intriguing scenario:

Over a 15-week semester, a graduate student has lunch in the campus food court every Tuesday, joined by various combinations of six friends. Throughout the semester, each friend joined him 11 times, each pair of friends joined him 9 times, each trio of friends 6 times, each quartet of friends 4 times, and each group of five friends 4 times. All seven individuals had lunch together once.

Using the principle of inclusion-exclusion, we deduce that there are no weeks where the graduate student dines alone.

However, I am curious about the practical possibility of this arrangement. Specifically, is it feasible to have 15 subsets $S_1, \dots, S_{15}$ of $[6]=\{1, 2, \dots, 6\}$ such that:

  • Each singleton subset of $[6]$ is contained in 11 of the $S_i$,
  • Each pair subset of $[6]$ is contained in 9 of the $S_i$,
  • And so forth...

Essentially, is the scenario as described possible, and are there sufficient or necessary conditions to validate its feasibility?

Update: The correct answer to the problem should be the student ate alone once. But that's not what I am asking.

NonalcoholicBeer
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  • I think you can understand this instead as having $6$ subsets $S_1, \ldots, S_6$ such that each $|S_i| = 11$, and each $|S_i \cap S_j| = 9$, and $|S_i \cap S_j \cap S_k| = 6$, etc. You can examine its feasibility using Venn diagram. – Vezen BU Feb 08 '24 at 05:24

2 Answers2

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You can use the Generalised inclusion-exclusion principle to determine the number $c_j$ of times that the student had lunch with exactly $j$ friends. With $b_k=15,11,9,6,4,4,1$ for $k=0,\ldots,6$ and

$$ c_j=\sum_{k=j}^6(-1)^{k-j}\binom kj\binom6kb_k $$

this yields

\begin{eqnarray*} c_0 &=& 15\binom00\binom60-11\binom10\binom61+9\binom20\binom62-6\binom30\binom63+4\binom40\binom64-4\binom50\binom65+1\binom60\binom66 \\ &=& 1\;, \end{eqnarray*}

so contrary to your conclusion, if this scenario had been possible, the student would have eaten alone exactly once. However,

\begin{eqnarray*} c_1 &=& 11\binom11\binom61-9\binom21\binom62+6\binom31\binom63-4\binom41\binom64+4\binom51\binom65-1\binom61\binom66 \\ &=& 30\;, \end{eqnarray*}

so the student would have had to have a one-on-one lunch with a single friend $30$ times. Since there are only $15$ weeks, we can conclude that this scenario is impossible. The negative culprits are

$$ c_2=9\binom22\binom62-6\binom32\binom63+4\binom42\binom64-4\binom52\binom65+1\binom62\binom66=-90 $$

and

$$ c_4=4\binom44\binom64-4\binom54\binom65+1\binom64\binom66=-45\;. $$

joriki
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Instead, if you consider the total set $[15]$ of the weeks, and consider $6$ subsets $S_1, \ldots, S_6 \subseteq [15]$ such that

  • $|S_i| = x_1 = 11, \forall i$
  • $|S_i \cap S_j| = x_2 = 9, \forall (i, j) \in \binom{[15]}{2}$
  • $|S_i \cap S_j \cap S_k| = x_3 = 6, \forall (i, j) \in \binom{[15]}{3}$
  • $|\bigcap_{i \in \mathcal{I}_4} S_i| = x_4 = 4, \forall \mathcal{I}_4 \in \binom{[15]}{4}$
  • $|\bigcap_{i \in \mathcal{I}_5} S_i| = x_5 = 4, \forall \mathcal{I}_5 \in \binom{[15]}{5}$
  • $|\bigcap_{i \in [6]} S_i| = x_6 = 1$

Then it is feasible iff it is feasible at each level

  • (Level 5) $|S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5| = |S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5 \cap S_6| + |(S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5 ) \setminus S_6| \Rightarrow |(S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5 ) \setminus S_6| = |S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5| - |S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5 \cap S_6|$, which requires $x_5 \geq x_6$; $x_5 - x_6$ is the number of occurrences for each group of size exactly $5$
  • (Level 4) $|S_1 \cap S_2 \cap S_3 \cap S_4| = |S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5 \cap S_6| + |(S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_5) \setminus S_6| + |(S_1 \cap S_2 \cap S_3 \cap S_4 \cap S_6) \setminus S_5| + |(S_1 \cap S_2 \cap S_3 \cap S_4) \setminus (S_5 \cup S_6)|$, which requires $x_4 \geq x_6 + 2 (x_5 - x_6) = 2 x_5 - x_6$; $x_4 - 2x_5 + x_6$ is the number of occurrences for each group of size exactly $4$
  • (Level 3) $x_3 \geq x_4 + (x_5 - x_6) + 2 (x_4 - 2x_5 + x_6) = 3x_4 - 3x_5 + x_6$; $x_3 - 3x_4 + 3x_5 - x_6$ is the number of occurrences for each group of size exactly $3$
  • (Level 2) $x_2 \geq x_3 + (x_5 - x_6) + 3 (x_4 - 2x_5 + x_6) + 3(x_3 - 3x_4 + 3x_5 - x_6)$; $x_2 - 4 x_3 + 6 x_4 - 4 x_5 + x_6$ is the number of occurrences for each group of size exactly $2$
  • (Level 1) $x_1 \geq x_2 + \binom{4}{1} (x_2 - 4 x_3 + 6 x_4 - 4 x_5 + x_6) + \binom{4}{2} (x_3 - 3x_4 + 3x_5 - x_6) + \binom{4}{3} (x_4 - 2x_5 + x_6) + \binom{4}{4} (x_5 - x_6)$
  • (Level 0) Using inclusion-exclusion to check $|\bigcup_i S_i| \leq 15$

The idea can be extended to other numbers of weeks and people.

Vezen BU
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