Grouping by Threes
Three committees are formed from eight people—F, G, H, I, J, K, L, M. Two of the committees have three members, and one of the committees has only two members.G serves with M.
L serves with only one other person.
F does not serve with M.
We start by symbolizing the conditions. The condition “G serves with M” is naturally symbolized as G = M. The condition “F does not serve with M” is symbolized as F ≠ M. The condition “L serves with only one other person” means that L is on the committee of two; we symbolize it as L = 2. The diagram will consist of three compartmentalized boxes. This gives the following schematic:
F G H I J K L M (H, I, J, K “wild”)
G = M
F ≠ M
L = 2
Com I 


Com II 


Com III 











L 

Before turning to the questions, two readily derived conditions should be noted. First, since G serves with M, and F does not serve with M, F cannot serve with G. Second, since L serves on the twoperson committee, L cannot serve with G or M (otherwise L would be on a threeperson committee).
Example1
Which one of the following is a committee?
 M, L, I
 G, F, M
 G, L
 G, H, I
 K, G, M
Solution
(A) is not a committee since L must serve on a committee of two.
(B) is not a committee since F cannot serve with G.
Neither (C) nor (D) is a committee since G and M must serve together.
Hence, by process of elimination, the answer is (E).
Example2
If F cannot serve with K, and K cannot serve with M, which one of the following must be false?
 F serves with L.
 F serves with J.
 L serves with H.
 H serves with I.
 I serves with M.
Solution
We shall use an indirect proof. Start with (A). If F serves with L, then G and M could serve on Committee I, K on Committee II, and the remaining people could serve at random without violating any initial condition. So F could serve with L. This eliminates (A). Next, test (B). If F serves with J on Committee I, then G and M would have to serve on Committee II. And the remaining people could be placed as follows:
This diagram does not violate any initial condition, so F could serve with J. This eliminates (B).
Next, test (C). There are two possible places for the pair G and M, Committee I and Committee II.
If G and M serve on Committee I, then F would have to serve on Committee II:
Clearly this diagram leaves no room for K since K cannot serve with either M or F.
The case with the pair G and M serving on Committee II leads to a similar result.
Hence, L cannot serve with H.
The answer is (C).
Com I  Com II  Com III  
F  J  H  G  M  I  L  K 
This diagram does not violate any initial condition, so F could serve with J. This eliminates (B).
Com I  Com II  Com III  
G  M  F  L  H 
Example3
If H serves with K, which one of the following cannot be true?
 F serves with K.
 J serves with F.
 I serves with M.
 F serves with L.
 J serves with L.
Solution
If H serves with K on Committee I, then G and M must serve on Committee II. (Why?) This gives the following diagram:
(The diagram with H and K on Committee II is not presented because it generates the same results.) Again we apply an indirect proof. Start with (A).
If F serves with K, then from the above diagram F must serve on Committee I. And we can place I and J on Committees II and III, respectively:
This diagram does not violate any initial condition, so F could serve with K. This eliminates (A).
Next, test (B). J and F cannot serve on Committee I, since from the above diagram H and K are already there.
Likewise, J and F cannot serve on Committees II and III.
Hence, the answer is (B).
Com I  Com II  Com III  
H  K  G  M  L 
Com I  Com II  Com III  
H  K  F  G  M  I  L  J 
Example4
If K, J, and I serve on different committees, which one of the following must be true?
 K serves with G.
 I serves on a committee of two.
 J serves on a committee of two
 H serves with F.
 J serves with F.
Solution
We shall construct counterexamples for four of the answerchoices; the one for which we cannot construct a counterexample will be the answer. Start with (A). Suppose K serves on Committee I and G serves on Committee II. Then from the condition G = M, we know that M must also serve on Committee II. And the remaining people can be placed without violating any initial condition as follows:
This diagram is a counterexample not only to (A) but to (C) and (E) as well. This eliminates (A), (C), and (E).
Next, test choice (B). Suppose that I serves on Committee I, with G and M. Then the remaining people can be grouped as follows:
This diagram does not violate any initial condition, so it is a counterexample to (B).
Hence, by process of elimination, the answer is (D).
Com I  Com II  Com III  
K  H  F  J  G  M  L  I 
Com I  Com II  Com III  
I  G  M  F  H  J  L  K 
Example5
Which one of the following conditions is inconsistent with the given conditions?
 K serves on a committee of three.
 M serves with H.
 M, H, and I serve together.
 F does not serve with G.
 H serves with L.
Solution
The first counterexample in Question 4 shows that K can serve on a threeperson committee. This eliminates (A).
Next, turning to choice (B), suppose M serves with H on Committee I. This forces G to also serve on Committee I.
Now place F on Committee III and the remaining people as follows:
This diagram does not violate any initial condition, so “M serves with H” is consistent with the initial conditions. This eliminates (B).
Next, turning to choice (C), suppose M, H, and I serve together on Committee I.
But since M must serve with G, there would then be four people on Committee I.
The same result occurs when M, H, and G are on Committee II.
Hence, (C) is inconsistent with the initial conditions, and the answer is (C).
Com I  Com II  Com III  
M  H  G  J  K  I  L  F 
This diagram does not violate any initial condition, so “M serves with H” is consistent with the initial conditions. This eliminates (B).