Bus Logic

A law school logic puzzle

Riders A, B, C, D, E, F, and G take buses 1, 2, and 3 to work. Use the rules below to determine which assignments are possible.

• Neither E nor G takes bus 1 on a day when B does.
• G does not take bus 2 on a day when D does.
• When A and F take the same bus, it is always bus 3.
• C always takes bus 3.

What the clues settle and what they do not

There is no single answer from these clues alone. They rule out incompatible combinations, but several riders can share a bus and no further condition fixes their choices. In fact, 406 assignments satisfy all four rules.

That is the central habit in logic games: distinguish between what must be true, what could be true, and what has not been determined yet. A complete puzzle would normally add more rules or ask a narrower question, such as which option could be true.

Counting the valid assignments

C is fixed on bus 3, so C contributes one choice. For A and F, there are 9 possible bus pairs, except for the two forbidden shared pairs on buses 1 and 2. That leaves 7 possibilities.

For B, D, E, and G, separate the cases by B’s bus. If B takes bus 1, E and G cannot take bus 1: this gives 10 assignments after also applying D and G’s bus 2 restriction. If B takes bus 2 or bus 3, E is unrestricted; each case gives 24 assignments. Together, that is 10 + 24 + 24 = 58 possibilities.

The total is therefore 1 × 7 × 58 = 406 valid assignments.

Turn prose into constraints

The first two rules forbid particular pairs: B and E cannot both be on bus 1; B and G cannot both be on bus 1; D and G cannot both be on bus 2. The third rule forbids A and F from sharing buses 1 or 2. C is fixed on bus 3.

Once written this way, each proposed arrangement can be checked systematically rather than guessed.