Math 184A Final Exam Solutions 7 Dec. 2005
1. (a) We’ll do this in three different ways.
(i) Count the number of rearrangements of 1,2,...,8 in two ways: First there are
8!. Second, seat people at a table, pick a side, and read off a list starting at that
side and going clockwise around the table. Thus 8! = (answer) ×4, giving 8!/4.
(ii) Designate a first person. Place that person on one side of the table (2 possible
seats) and then arrange the remaining 7 people, giving 2 ×7!.
(iii) Use Burnside’s Lemma. The rotations are 0◦, 90◦, 180◦and 270◦.N(0◦) = 8!
and, since all the people are different, N(r◦) = 0 for r6= 0. Thus we have
1
4(8! + 0 + 0 + 0 + 0).
(b) This could be done in various ways. The easiest is to use Burnside’s Lemma.
Call positions around the table 1,2,...,8 reading clockwise as shown in the first
picture in the exam problem. The group elements in cycle form are
no rotation: (1)(2)(3)(4)(5)(6)(7)(8) 90◦rotation: (1,3,5,7)(2,4,6,8)
180◦rotation: (1,5)(3,7)(2,6)(4,8) 270◦rotation: (1,7,5,3)(2,8,6,4)
Since chairs must be the same (color) on a cycle, we choose which cycles should
have red chairs, getting the answer
1
48
4+2
1+4
2+2
1 =70 + 2 + 6 + 2
4= 20.
The other way is to attempt to list all 20 solutions, but it is very easy to omit a
solution or count it twice.
2. (a) There are N=n
2possible edges. Since we must choose qof them, the answer
is N
q.
(b) Since the vertices in Scannot be used, there are M=n−|S|
2possible edges and
the answer is M
q.
(c) Use (b) and the Principle of Inclusion and Exclusion.
3. There is no such graph. Suppose Gwere such a graph. We can construct a spanning
tree Tfor Gby removing edges one at a time. Using subscripts to indicate the number
of edges, we obtain the sequence G=G25, G24, G23 , G22, G21 , G20, G19 =Tsince a
tree has one less edge than it has vertices. Thus we removed six edges. When an edge
eiis removed from Gi, it must belong to at least one cycle of Gi(since otherwise Gi−1
would not be connected). Thus, removing eidestroys at least one cycle of Giand
hence of G. Since we removed six edges, we must have destroyed at least six cycles of
G. Thus Gmust have at least six cycles.
4. Apply Principle 11.7 with F(x, y) = x(ey−y)−y. Thus we must solve the pair of
equations
F(r, s) = r(es−s)−s= 0 and Fy(r, s) = r(es−1) −1 = 0.
