Question

Now replace \(M\) by another \(O\) (so the multiset is \(\{E,H,O,O,R,T\}\)). Which of the following \emph{must be false?}

• A. O's can be on opposite walls in the middle
• B. Either R or H will be next to an O
• C. Either R or H will be next to either T or E
• D. T will be opposite to either O or E

Hint:

When one colour cannot sit next to certain colours, putting that colour in the \emph{middle} forces two legal neighbours; check if you have enough legal letters to supply both sides on \emph{both} walls.

Answer 1

The Correct Option is A

Step 1: Try to realize each claim.
\underline{(a)} If both centres are \(O\) (i.e., \(E_2=O\) and \(W_2=O\)), then on each wall the centre \(O\) must have two neighbours that are \emph{not} \(E\) or \(T\). But the only remaining letters not banned next to \(O\) are \(R\) and \(H\); across the two walls we would need \emph{four} side neighbours from \(\{R,H\}\) while only two such letters exist—impossible. Therefore (a) \emph{cannot} be true \(\Rightarrow\) it \emph{must be false}. \smallskip \underline{(b)} With two \(O\)'s and the bans \(O\!-\!E\), \(O\!-\!T\), at least one of \(R\) or \(H\) must sit next to an \(O\) (indeed, typically both) — \emph{must be true}. \smallskip \underline{(c)} Place walls as, e.g., \(O\!-\!R\!-\!O\) and \(E\!-\!T\!-\!H\); here \(H\) is next to \(T\) (or \(E\) depending on the order). So (c) is realizable — not necessarily false. \smallskip \underline{(d)} It is \emph{not forced} (e.g., with the previous layout, put \(E\)–\(T\)–\(H\) opposite \(O\)–\(R\)–\(O\) so \(T\) faces \(R\)); hence (d) is not "must be false". Step 2: Conclude.
Only (a) is impossible under the doubled-\(O\) constraint. \[ \boxed{\text{(a)}} \]