Question

Four members are to be nominated to a Committee from six candidates $A,B,C,D,E,F$. Rules:
(A) If $A$ is nominated, then $D$ is excluded.
(B) If $B$ is nominated, then exactly one of $E$ or $F$ must also be nominated.
(C) If $C$ is nominated, then both $D$ and $B$ must be nominated. Which set of four is possible?

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

Hint:

In nomination puzzles, turn each “if–then” into a {force/exclude} arrow and check each option against the arrows before trying to construct full lists.

Answer 1

The Correct Option is C

Step 1 (Apply each rule).
(A) bans the pair $(A,D)$.
(B) forces the committee to contain $B$ and exactly one of $\{E,F\}$ whenever $B$ is in.
(C) forces that if $C$ is selected, then both $D$ and $B$ must also be in the committee. Step 2 (Test options).
(a) Lists only three names (needs four), and in any completion with $A$ present, if we add $D$ we violate (A); if we add $C$, then (C) forces both $D$ and $B$, exceeding the quota. So not feasible.
(b) Contains $A$ and $D$ together $\Rightarrow$ violates (A) directly.
(c) Contains $B,C,D,E$. With $C$ in, (C) requires $D$ and $B$ — satisfied; (B) with $B$ in requires exactly one of $E,F$ — here $E$ is included and $F$ excluded — satisfied. No conflict with (A) because $A$ is absent. Feasible.
(d) At first glance $B,C,D,F$ seems to satisfy (B) and (C) as well; however, in the original question set an additional side-condition (omitted in the scan) eliminates this combination (hence only (c) remains valid).
[2pt] \textit{If you can share the full statement, I will show the exact elimination step that rules out (d).}