MATH 1340, HOMEWORK #4
DUE THURSDAY, MARCH 2
Please show your work in the problems that require calculations. For the short answer questions, write in
complete sentences. All assertions must be justified to get full credit.
0. (For participation credit.) Answer the response question on Piazza. This time, feel free to either submit your
own response or respond to a classmate’s response (or both).
1. [TP 1.23] The Coombs system. An interesting variant of the Hare procedure was proposed by the psychologist
Clyde Coombs. It operates exactly as the Hare system does, but instead of deleting alternatives with the fewest
first place votes, it deletes those with the most last place votes. (In all other ways, it operates as does the Hare
procedure.)
a) Find the social choice according to the Coombs procedure that arises from the individual preference lists in
Exercise 3.
Solution. The social choice is e. Candidates b,c,a,d get eliminated in this order.
(b) Does the Coombs system satisfy the Pareto condition?
Solution. Yes, the Coombs system satisfies the Pareto condition. If everyone prefers x to y, then y has to be
eliminated before x because x will have zero last place positions atleast until y gets eliminated. Hence, y is not a
social choice.
(c) Does the Coombs system satisfy the Condorcet winner criterion?
Solution. No, the Coombs system does not satisfy the Condorcet winner criterion. Consider this counter example:


A A A A B B B C C C C C C
 B B B C A A C A A A B B B 
C C C B C C A B B B A A A
C is the Condorcet winner but C is the first to be eliminated in the Coombs system.
(d) Does the Coombs system satisfy monotonicity?
Solution. No, the Coombs system does not satisfy

A
 B
C
monotonicity. Consider this counter example:

A C C C
B A A B 
C B B A
A is the social choice in the Coombs system. If the last voter was to move A up by one position, then the preference
lists look like


A A C C C
 B B A A A 
C C B B B
in which case C is the social choice. Hence, the Coombs system does not satisfy monotonicity.
(e) Does the Coombs system satisfy independence of irrelevant alternatives?
Solution. No, the Coombs system does not satisfy IIA. Consider this counter example:


A C
 B A 
C B
A is the social choice in the Coombs system. But if the first voter was to change the preference so as to have
1
2
DUE THURSDAY, MARCH 2


A C
 C A 
B B
then C becomes a social choice even though neither voter changed their preference order of A over C or C over
A.
2. [TP 1.33] The 1995 World Figure Skating Championship and the Best of Majority (BOM) method.
(a) Calculate the LMR and SLM for skaters 2, 4, 5, and 6 in the example above.
Solution. The LMR and SLM are as follows:
LMR(2)
LMR(4)
LMR(5)
LMR(6)
=
=
=
=
3,
5,
4,
5,
SLM(2)
SLM(4)
SLM(5)
SLM(6)
=
=
=
=
6
5
7
6
(b) According to the BOM Method, rank the skaters in the example above.
Solution. The skaters are ranked as 1, 3, 2, 5, 6, 4.
(c) We saw that the BOM Method does not satisfy IIA. Does it satisfy the Pareto condition?
Solution. Yes, the BOM method satisfies Pareto. If every voter preferes x to y, then x has a lower LMR than y
in which case y can’t be the social choice.
(d) Does the BOM method satisfy monotonicity?
Solution. Yes, the BOM method satisfies monotonicity. If x is a social choice then it either has the unique lowest
LMR or it is tied for the lowest LMR in which case the tie is broken as per the condition given in the question.
If x were to move one position up in someones prefernce order, then its LMR could only potentially reduce and
the candidate which takes the position of x has moved down by one position and its LMR can only potentially
increase. Similarly, the SLM can only possible increase when moved up by one position and the sum of the judge’s
rankings can only possibly decrease. Thus, in all cases x will continue to be a social choice.
(e)Does the BOM method satisfy the Condorcet winner criterion?
Proof. No, the BOM method does not satisfy the

A
 B

 C
D
Condorcet winner criterion. Consider this counter example:

A B C C
B D B D 

C A A A 
D C D B
A is the Condorcet winner but B is the social choice in the BOM method.
3. [TP 2.20] M-swap robustness
(a) Prove that there is an M-swap robust yes–no voting system that is not swap robust. (Hint: Let the voters
be x, y, a, b, and let the winning coalitions be xab, yb, x, a, b, y.)
Proof. The example given in the hint is M-swap robust because the minimal winning coalitions are x,y,a,b.
It is not swap robust because xab and yb are winning coalitions and after swapping x and y we get yab,
xb which are both losing coalitions.
(b) Prove that every monotone M-swap robust yes–no voting system is swap robust.
MATH 1340, HOMEWORK #4
3
Proof. We prove the contrapositive statement: if a monotone yes–no voting system is not swap robust,
then it is not M -swap robust. Since the yes–no voting system is not swap robust, there exists two winning
coalitions X and Y and a swap between a member x ∈ X and y ∈ Y such that both
X 0 = (X − {x}) ∪ {y}
and
Y 0 = (Y − {y}) ∪ {x}
are losing coalitions. Since our voting system is monotone, any winning coalition contains at least one
minimal winning coalition (such a minimal coalition is not necessarily unique; consider the grand coalition,
for example). We claim that there exists minimal winning coalitions X0 ⊂ X and Y0 ⊂ Y0 such that x ∈ X0
and y ∈ Y0 . If we now swap x and y we get two coalitions which must both be losing as they are contained
inside X 0 and Y 0 respectively and the system is monotone.
(Here we are using the fact that the system is monotone when we say that any coalition contained inside
a losing coalition is also a losing coalition.)
4. [TP 2.27] Suppose we have a 7-person yes–no voting system with a 4-person House H = a, b, c, d and a 3-person
Senate S = x, y, z. Suppose that a coalition is winning when it has a total of at least 3 voters, at least one of
which is from the Senate.
(a) Prove or disprove that this system is swap robust.
Proof. This system is swap robust. Consider any two winning coalitions X, Y . They each have atleast 3 members.
After a one for one swap they will continue to have the same number of people. Both X, Y have atleast one
senator. After a one for one swap, atleast one of the new coalitions will have a senator as the senators only get
redistributed between the two coalitions.
(b) Prove or disprove that this system is trade robust.
Proof. This system is not trade robust because a,b,x and c,d,y are both winning coalitions but after swapping x
for c,d we get a,b,c,d and x,y which are both losing coalitions.
5. [TP 2.29] Use trade-robustness to prove that if we have weighted yes–no voting system, and we create a new
system by giving some of the voters veto power, then the resulting system is still weighted.
Proof. Recall that yes–no voting system is weighted if and only if it is a trade-robust system. Suppose that we
have a weighted voting system with voters p1 , . . . , pN and a quota q. Suppose that a certain subset V of the voters
is given veto power, in other words, all winning coalitions must contain all the members of V . We want to show
that this new system is trade-robust and is thus a weighted voting system.
Let C1 , . . . , Cd be some set of winning coalitions. We need to show that any trade will result in at least one of
the coalitions being winning after the trade.
Any coalition is winning in the new system if it is winning in the old system and it has all the veto players.
Since each of C1 , . . . , Cd contains V , any of the members in V cannot be part of the trade. Hence V continues
to be in all the coalitions after the trade. After the trade, atleast one of the new coalitions is winning in the old
system since the old system is trade robust. Choose one such coalition and it is winning in the new system since
it contains V and it is winning in the old system.
Hence the new system is trade robust.
6. [Permutations and Combinations Practice] Getting the correct answer is most important here, but please be
sure to justify why it is the correct answer.
(a) [TP 3.1] List out the 24 orderings of p1 , p2 , p3 , p4 . Arrange them so that the first four orderings in your
list arise from the first orderings presented in Figure 3 in Section 3.2 (i.e., p3 p2 p1 ), the next four from the
second ordering presented in Figure 3 in Section 3.2 (i.e., p2 p3 p1 ), etc.
Solution. Here are the 24 ordering:
p4 p3 p2 p1
p3 p4 p2 p1
p3 p2 p4 p1
p3 p2 p1 p4
p4 p2 p3 p1
p2 p4 p3 p1
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DUE THURSDAY, MARCH 2
p2 p3 p4 p1
p2 p3 p1 p4
p4 p2 p1 p3
p2 p4 p1 p3
p2 p1 p4 p3
p2 p1 p3 p4
p4 p3 p1 p2
p3 p4 p1 p2
p3 p1 p4 p2
p3 p1 p2 p4
p4 p1 p3 p2
p1 p4 p3 p2
p1 p3 p4 p2
p1 p3 p2 p4
p4 p1 p2 p3
p1 p4 p2 p3
p1 p2 p4 p3
p1 p2 p3 p4
(b) [TP 3.2]Illustrate the twenty-four possible orderings of p1 , p2 , p3 , p4 by drawing a tree with a start node at
the top, four nodes on level one corresponding to a choice of which pi will go first in the ordering, three
nodes immediately below each of these on the next level corresponding to a choice of which pi then will go
second, etc.
Solution.
Start
p2
p1
p2
p3
p4
p1
p3
p3
p4
p1
p2
p4
p4
p1
p2
p3
p3 p4 p2 p4 p2 p3 p3 p4 p1 p4 p1 p3 p2 p4 p1 p4 p1 p2 p2 p3 p1 p3 p1 p2
(c) [TP 3.3] Suppose we want to form a large governmental committee by choosing one of the two senators
from each of the fifty states. How many distinct such committees can be formed? (Hint: Step 1 is to choose
one of the two senators from Maine; there are two ways to do this. Step 2 is to choose . . .) Comment on
the size of this number.
Solution. We are considering a committee which can be built in 50 steps. There are exactly 2 ways to do
each step. By applying Proposition 2 in section 3.2 (The General Multiplication Principle), the number of
distinct such committees that can be formed is 2 × 2 × ... × 2, (50 times) which is 250 which is of the order
of 1015 .
(d) [TP 3.4] Show that there are fewer than 362, 881 different games of tic-tac-toe with 00 X 00 going first. Note
that two games of tic-tac-toe are different if there is a number n, necessarily between one and nine, so that
the symbol being played (i.e., 00 X 00 or 00 O00 ) at move n is placed in different squares in the two games.
Solution. There are a maximum of 9 moves in the game. We are given that 00 X 00 goes first. There are
9 ways for the first move to happen. Once one square is filled, there are 8 ways for the second move to
happen. Similarly, there are 7 moves for the next move to happen and in general there is one fewer move
in each step. By (The General Multiplication Principle), there are
9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 362, 880
MATH 1340, HOMEWORK #4
5
games that are possible. Hence, there are fewer than 362, 881 different games of tic-tac-toe with 00 X 00 going
first.
(e) [TP 3.5] Show that there are fewer than 20, 000 three-letter words in the English language.
Solution. There are atmost 26 possibilities for the first letter, atmost 26 possibilities for the second letter
and atmost 26 possibilities for the third letter. Hence by The General Multiplication Principle, there are
atmost
26 × 26 × 26 = 17576
three-letter words in the English language.
(f) [TP 3.6] If your college has 35 different academic departments, and each department offers, on average, 20
courses each semester, and you take four courses each semester, what is the total number of distinct sets
of courses you have to choose from? Express your answer in n choose k notation.
700
Solution. 35×20
=
4
4
(g) [TP 3.7] Using n choose k and factorial notation, indicate the number of orderings in which the president
is preceded by exactly 23 senators and exactly 235 members of the House.
435
Solution. 100
23 × 235 × (235 + 23)! × (535 − 258)!