Which one of the following, if substituted for the condition that Molina must be assigned to exactly one committee, w...

Irina on September 1, 2019

@Tony,

Let's quickly look at the setup:

5 volunteers - H J K M N are assigned to X Y Z.

Each committee will have 3 volunteers assigned to it with each volunteer holding one of the three positions: leader, secretary, or treasurer.

We thus have 5 people to fill in 9 slots.

X ___ ___ ___
Y ___ ___ ___
Z ___ ___ ___
L S T

The following conditions apply:

(1) N -> N (L)

(2) M = 1

(3) K must be assigned to Y but cannot be assigned to Z.

X ___ ___ ___
Y ___ ___ ___ K
Z ___ ___ ___ ~K
L S T

(4) J must be the secretary for Y but cannot be assigned to X or Z

X ___ ___ ___ ~J
Y ___ J ___ K
Z ___ ___ ___ ~K ~J
L S T

These rules tell us that out of 9 slots, 1 of them must be filled with J, one with M, and two with K per rule (3) - K must be assigned to Y but cannot be assigned to Z. We still have 5 slots to fill and only 2 volunteers left - N & H, meaning they must either be assigned to 3 or 2 committees each.

Note that we cannot have K assigned only to Y because we would not be able to fill the rest of the roles as N & H cannot both be assigned to all 3 committees due to J & K already being assigned to committee Y. Thus, we could only have either H or J on committee Y, but not both.

The number of potential positions each volunteer could occupy could thus be summarized as:

M =1
J =1
K = 2
N = 3 or 2
H = 3 or 2

The question asks us what rule would have the same effect as M=1?

The rule must lead to the same inference as in our initial setup that M could only take on one position. Let's look at the answer choices:

(A) H must be assigned to more committees than M

Incorrect. This rule only allows us to infer to M could be assigned to 2 committees at most, not 1.

(B) J must be assigned to more committees than M.

Incorrect. This rule would result in M being assigned to 0 committees because J is assigned to 1, which is impossible as everyone must be assigned to at least 1 committee to fill all the positions.

(C) K must be assigned to more committees than M.

Correct. We can see that K must be assigned to 2, and this rule would mean M has to be assigned to 1.

(D) M must be assigned to more committee than H.

Incorrect. This would lead to H being assigned to 0 committees, which is impossible.

(E) N must be assigned to more committees than M.

Incorrect. This rule only allows us to infer to M could be assigned to 2 committees at most, not 1.

Does this make sense?

Let me know if you have any further questions.