Question

A tourist must tour {four} places out of $A,B,C,D,E,F,G$. At most two can be {holiday} places and at least two must be {business} places. Types: $A,B,C,D$ are Business; $E,F,G$ are Holiday. Rules: (a) If $A$ is included, then both $C$ and $G$ are excluded. (b) If neither $E$ nor $F$ is included, then $B$ or $G$ or both may be included. (c) If $G$ is included, then $D$ cannot be included. Which combination is {possible}?

• A. $A, C, E$ and $F$
• B. $B, G$ and $E$
• C. $A, D$ and $G$
• D. $A, B$ and $D$

Hint:

Under selection constraints, try to {complete} a listed partial set to a full valid set. Contradictions (forced exclusions) quickly rule choices out.

Answer 1

The Correct Option is D

Step 1 (Decode). We must be able to extend the listed combination to a valid set of {four} with $\le 2$ holidays and $\ge 2$ business places while obeying (a)–(c). Step 2 (Check each option).
(a) Contains $A$ and $C$ together $\Rightarrow$ violates (a) directly $\Rightarrow$ impossible.
(c) Contains $G$ and $D$ together $\Rightarrow$ violates (c) $\Rightarrow$ impossible.
(b) Already uses two holidays ($E$ and $G$). The fourth place must be {business}. It cannot be $D$ (forbidden with $G$ by (c)), and cannot be $A$ (forbidden with $G$ by (a) since $A$ would also force out $G$). The only admissible choice is $C$, but (b) as stated gives us no certainty that $C$ is included; the listed combination by itself is thus not guaranteed to be extendable as required.
(d) $A,B,D$ are all {business}. With $A$ present, simply avoid $C$ and $G$ per (a); choose any one of $E$ or $F$ as the fourth (holiday) place. Example valid tour: $\{A,B,D,E\}$ satisfies “at most $2$ holidays” and “at least $2$ business” and respects (a)–(c). Hence (d) is possible.