Solutions to Homework 6
Prof. Silvia Fernández
Discrete Mathematics
Math 513A, Fall 2008
1. [3.8 #2] We verify that in Figure 3.96(b) there is no K3 and no I5 . Assume that there
is a K3 . Call it G0 . By symmetry we can assume that vertex 1 is in G0 . Since the only
neighbors of 1 are 2, 6, 9, and 13 then two of these four vertices must be in G0 and thus they
must be adjacent. But the graph induced by {2, 6, 9, 13} is an independent graph I4 getting
a contradiction.
Assume now that there is an I5 , call it G00 . Then there are no consecutive vertices (13 and
1 are considered consecutive, around the polygon) in G00 . By rotational symmetry we can
assume that the vertices of G00 are 1, 3, 5, 7, 9, 11. But {1, 9} is an edge and thus G00 6= I5 .
2. [3.8 #5] (a,b) The graph induced by {a, c, e} is a clique of 3 vertices.
(c,e) There is no clique or independent set of 3 vertices.
(d) The graph induced by {a, b, c} is an independent set of 3 vertices.
(f) The graph induced by {a, c, d} is a clique of 3 vertices.
3. [3.8 #10] (a) There are exactly 16 edges incident in a. Each edge gets one of the 3 available
colors. By the Pigeonhole Principle at least d16/3e = 6 edges must have the same color.
(b) Let G0 be the subgraph induced by {b, c, d, e, f, g}. Then G0 is a K6 whose edges are
colored red, blue or white. If one of these edges is red then together with a it forms a red
triangle. If none of the edges of G0 is red then G0 is a K6 whose edges are 2-colored (blue or
white). Since R(3, 3) = 6, G0 must have a blue or white triangle.
4. [3.8 #11a, d] (a) Consider a coloring of the edges of K3 with 2 colors. By the Pigeonhole
Principle at least 2 edges get the same color. Since these two edges must share a vertex, they
form a monochromatic L3 .
(b) We prove that R (L3 , Z4 ) = 4. In other words, the following two statements are satisfied.
First. There is a (red,blue)-coloring of the edge of K3 with no red L3 and no blue Z4 .
Color all three edges of K3 blue.
Second. Any (red,blue)-coloring of the edges of K4 contains a red L3 or a blue Z4 .
Suppose by contradiction that there is such a coloring. This coloring should contain at least
one red edge (otherwise we have a blue K4 , which obviously contains a Z4 ). Say uv is a red
edge and the other two vertices are x and y. Note that no two red edges are incident (this
would form a red L3 ). Thus ux, xv, vy, and yu must be blue. But these for edges form a Z4
(namely, uxvyu) getting a contradiction.
5. [7.1.13] Use Inclusion and Exclusion Principle where a1 , a2 , a3 , and a4 are the properties of
being divisible by 2, 3, 5, and 7 respectively. Then the number we are looking for is
¥ ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦
− 3 − 5 − 7 + 6 + 10 + 14
N [a01 , a02 , a03 , a04 ] = 600 − 600
2
¥ ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦ ¥ 600 ¦
+ 600
15 + 21 + 35 − 30 − 42 − 70 − 105 + 210 = 137.
6. [7.1.17] Use Inclusion and Exclusion Principle where a permutation of {1, 2, 3, 4, 5, 6} has
property ai if i + 1 immediately follows i, for 1 ≤ i ≤ 5. That is, a permutation has property
a1 , a2 , a3 , a4 , or a5 if it has 12, 23, 34, 45, or 56 respectively. For example the permutation
345621 satisfies properties a4 and a5 because 45 and 56 appear.
1
Then the number we are looking for is N [a01 , a02 , a03 , a04 ]. We claim that N [ai ] = 5!, N [ai , aj ] =
4!, N [ai , aj , ak ] = 3!, N [ai , aj , ak , al ] = 2!, N [a1 , a2 , a3 , a4 , a5 ] = 1.and thus
µ ¶
µ ¶
µ ¶
µ ¶
µ ¶
£ 0 0 0 0¤
5
5
5
5
5
5! +
4! −
3! +
2! −
1! = 309.
N a1 , a2 , a3 , a4 = 6! −
1
2
3
4
5
Note that N [a1 ], for example, counts the number of permutations of {1, 2, 3, 4, 5, 6} where
two immediately follows 1 (having 12). This is the same as the number of permutations
of {12, 3, 4, 5, 6} (here 12 is a single element). Similarly to evaluate N [a1 , a2 ] we count the
permutations of the 4-element set {123, 4, 5, 6} because any permutation having 12 and 23
actually has 123. The analysis for the pairs [a2 , a3 ] , [a3 , a4 ] , and [a4 , a5 ] is the same. All
other pairs behave differently but still give 4!. For example to evaluate N [a1 a3 ] look at the
permutations of the 4-element set {12, 34, 5, 6}.
N [ai , aj , ak ] is always 3! but we have three different types of triples.
To evaluate triples like
N [a1 , a2 , a3 ]
N [a1 , a2 , a4 ]
N [a1 , a3 , a5 ]
count permutations of
{1234, 5, 6}
{123, 45, 6}
{12, 34, 56} .
Finally
To evaluate
N [a1 , a2 , a3 , a4 ]
N [a1 , a2 , a4 , a5 ]
N [a1 , a3 , a4 , a5 ]
N [a2 , a3 , a4 , a5 ]
count permutations of
{12345, 6}
{123, 456}
{12, 3456}
{1, 23456}
and the only permutation satisfying all properties is 123456.
7. [7.1.18c, 19c] We compute the chromatic polynomial of K4 in two different ways.
First. Using Inclusion and Exclusion Principle.
Label the vertices and edges of K4 as in the following picture.
1
a
b
2
5
3
d
4
6
c
For 1 ≤ i ≤ 6 let ai be the property that in a vertex coloring of K4
the vertices forming edge i get the same color.
Then N [ai ] = x½3 , N [ai , aj ] = x2 ,
x2 if ai , aj , ak form a triangle (4 possibilities)
N [ai , aj , ak ] =
x otherwise,
N [any other set of properties] = x.
£
¤
Then P (K4 , x) = N a01 , a02 , a03 , a04 , a05 , a06
µ ¶
¶
µ ¶
µ ¶
µµ ¶
6
6
6 2
6
4
3
2
x+x
x−
x − 4x −
−4 x+
= x − 6x +
5
4
2
3
= x4 − 6x3 + 11x2 − 6x.
Second. Using Theorem 7.3.
The following are all induced subgraphs of K4 together with their number of edges e and
number of components c.
2
e=0
c =4
one
e=2
c=2
three
e=1
c =3
s ix
e=2
c=2
twelve
e=3
c=2
four
e=3
c=1
twelve
e=6
c=1
one
e=4
c=1
three
e=5
c =1
s ix
e=4
c=1
twelve
e=3
c=1
four
So h (0, 4) = 1, h (1, 3) = 6, h (2, 2) = 15, h (3, 2) = 4, h (3, 1) = 16, h (4, 1) = 15, h (5, 1) =
6, h (6, 1) = 1 and all other h (e, c) = 0. Then by Theorem 7.3 we have
X
(−1)e h (e, c) xc
P (K4 , x) =
e,c
4
= x − 6x3 + 15x2 − 4x2 − 16x + 15x − 6x + x
= x4 − 6x3 + 11x2 − 6x.
8. [7.2.6(modified)] How many words of length 6 have an even number of different vowels?
Solution. Let A be the set of all words of length 6. Then N = |A| = 266 .
Consider the following 5 properties on A.
a1 = NOT having the letter a,
a2 = NOT having the letter e,
a3 = NOT having the letter i,
a4 = NOT having the letter o,
a5 = NOT having the letter u,
Then (using the notation in chapter 7)
s1 =
s2 =
s3 =
s4 =
s5 =
µ ¶
5
ΣN (ai ) =
256 = 5 · 256 ,
1
µ ¶
5
ΣN (ai , aj ) =
246 = 10 · 246 ,
2
µ ¶
5
ΣN (ai , aj , ak ) =
236 = 10 · 236 ,
3
µ ¶
5
226 = 5 · 226 ,
ΣN (ai , aj , ak , al ) =
4
µ ¶
5
ΣN (a1 , a2 , a3 , a4 , a5 ) =
216 = 216 .
5
By definition e1 = number of words in A satisfying EXACTLY one property. For example, any
word satisfying property a1 but not properties a2 , a3 , a4 , a5 . This means means NOT HAVING the
letter a but HAVING the letters e, i, o, u . That is,
e1 = # words in A having EXACTLY 4 vowels.
3
Similarly
e3 = # words in A having EXACTLY 2 vowels,
e5 = # words in A having EXACTLY 0 vowels.
Using Theorem 7.5, the number of words in A having an even number of different vowels is
!
Ã
5
X
1
e1 + e3 + e5 =
(−2)t st
s0 −
2
t=0
¢
1¡ 6
=
26 − 266 + 2 · 5 · 256 − 4 · 10 · 246 + 8 · 10 · 236 − 16 · 5 · 226 + 32 · 216
2
= 157 140 941.
4