A committee of exactly five members must be chosen from a group of seven candidates: three men (M1, M2, M3) and four women (W1, W2, W3, W4). The selection must adhere to the following conditions: 1. At least two men must be selected. 2. If W1 is selected, then W3 must also be selected. 3. M2 and W4 cannot both be selected. 4. If M3 is selected, then W2 cannot be selected. Which of the following is a pair of candidates that CANNOT both be selected for the committee?

Correct answer: B. M3 and W1

If M3 and W1 are both selected, then by rule 2, W3 must also be selected. This means M3, W1, W3 are selected. By rule 4, if M3 is selected, W2 cannot be selected. So, W2 is out. We need 5 members, so far M3, W1, W3 are in. We need 2 more. The remaining candidates are M1, M2, W4. We cannot select W4 with M2 (rule 3). If we select M2, then we can't select W4. If we don't select M2, we must select M1 and W4. But M2 is still available. If M3, W1, W3 are in, W2 is out. We need 2 more from M1, M2, W4. We must have at least two men (rule 1). M3 is one man. We need one more. If we pick M2, we cannot pick W4 (rule 3). So we'd have M3, W1, W3, M2, M1 (5 members). This is a valid committee. Wait, let's re-evaluate. If M3 and W1 are both selected: M3, W1. Rule 2: W3 must be selected. So M3, W1, W3. Rule 4: if M3 selected, W2 cannot be selected. So M3, W1, W3, W2 (out). We have 3 members. We need 2 more from M1, M2, W4. We need at least 2 men. M3 is one man. So we need M1 or M2. If we pick M2, we cannot pick W4 (rule 3). So we would have M3, W1, W3, M2. We need one more. Only M1 is left. So M3, W1, W3, M2, M1. This is a valid committee. Let's re-read the question. "CANNOT both be selected". My example shows they CAN. Let's re-examine rule 4. M3 selected => W2 NOT selected. Okay. M2 and W4 cannot BOTH be selected. Okay. W1 selected => W3 selected. Okay. At least two men. Okay. Let's try to construct a scenario where M3 and W1 are *both* selected. The committee would include M3, W1, W3 (from rule 2). Due to M3, W2 is NOT selected (from rule 4). We have 3 members: M3, W1, W3. We need 2 more members from the remaining candidates: M1, M2, W4. We need at least two men; we currently have M3. So we need at least one more man (M1 or M2). If we pick M2, then W4 CANNOT be picked (rule 3). So if we pick M2, we must pick M1 to fill the 5th spot. Committee: M3, W1, W3, M2, M1. This is valid. So M3 and W1 CAN both be selected. This means C is not the answer. Let's check D. If W2 and W4 are both selected. We need 5 members. W2, W4 are IN. Rule 3: M2 and W4 cannot both be selected. So M2 is OUT. We have W2, W4. We need 3 more. From M1, M3, W1, W3. Rule 1: at least 2 men. So we need M1 and M3. If M3 is selected, then W2 CANNOT be selected (rule 4). But we already assumed W2 is selected. This is a contradiction. Therefore, W2 and W4 cannot both be selected. This is the correct answer. My initial assessment of C was flawed due to a misstep in constructing the committee. The contradiction with W2 and W4 is clear.