Discussion
Consider a set \(U\) of 23 different compounds in a Chemistry lab. There is a subset \(S\) of \(U\) of 9 compounds, each of which reacts with exactly 3 compounds of \(U\). Consider the following statements:
I. Each compound in \(U \setminus S\) reacts with an odd number of compounds.
II. At least one compound in \(U \setminus S\) reacts with an odd number of compounds.
III. Each compound in \(U \setminus S\) reacts with an even number of compounds.
Which one of the above statements is ALWAYS TRUE?
Answer: B
The statement that is always true is only II: at least one compound in \(U \setminus S\) reacts with an odd number of compounds.
Model the compounds as vertices of a graph, with an edge for each reacting pair. The handshaking lemma says the degrees add up to twice the number of edges, an even number.
The 9 compounds in \(S\) contribute \(9 \times 3 = 27\), which is odd. So the 14 compounds in \(U \setminus S\) must contribute an odd total, and a sum of integers is odd only if at least one of them is odd.
Statement I can fail (only one compound of \(U \setminus S\) needs odd degree), and III always fails, since if every degree in \(U \setminus S\) were even, the total would be odd.
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