Hints and Solutions to Selected Exercises

Abstract

Abstract

1.1

Let ${\mathcal P}_{O}$ be the class of partitions into odd number of parts. Its symbolic description is

$$\displaystyle \begin{aligned} {\mathcal P}_{O}=\text{Seq}(\{1\})\times \text{ Seq}(\{3\}) \times \text{ Seq}(\{5\})\times \cdots \end{aligned}$$

On the other hand, the class ${\mathcal P}_d$ of distinct partitions can be written as

$$\displaystyle \begin{aligned} {\mathcal P}_d=(\epsilon +\{1\})\times (\epsilon+\{2\})\times \cdots \end{aligned}$$

1.2 A path in ${\mathcal D}_k$ is a Dyck path followed by a $\searrow$ step, followed by a second Dyck path and $\searrow$ step and so on up to k times and finally an additional Dyck path:

$$\displaystyle \begin{aligned} {\mathcal D}_k=({\mathcal D}_0\times \{\searrow\})^k\times {\mathcal D_0}. \end{aligned}$$

From the above description one gets

$$\displaystyle \begin{aligned} D_k(z)=z^kD_0^{k+1}(z), \end{aligned}$$

For $k=2$, the 9 paths of length 6 in ${\mathcal D}_2$ are depicted in Fig. 10.1.

Fig. 10.1

${\mathcal D}_2$ paths

1.3

$$\displaystyle \begin{aligned} {\mathcal S}=\{\epsilon\}+\{\rightarrow\rightarrow\}\times {\mathcal S}+\{\nearrow\}\times {\mathcal S} \times \{\searrow\} \times {\mathcal S}. \end{aligned}$$

The length of the path must be even, so if we declare the size of a path of length 2n to be n we get the functional equation

$$\displaystyle \begin{aligned} S(z)=1+z(S(z)+S^2(z)). \end{aligned}$$

1.4

As in the case of Dyck paths, in order to find a recursive formula we consider the smallest $i>0$ such that the path contains the point $(i,0)$. Every path in ${\mathcal D}^{\leqslant m}$ can be described as a step up followed by a path in ${\mathcal D}^{m-1}$ followed by a step down and a path in ${\mathcal D}^{\leqslant m}$:

$$\displaystyle \begin{aligned} {\mathcal D}^{\leqslant m}=\epsilon +\{\nearrow\}\times {\mathcal D}^{\leqslant m-1}\times \{\searrow\}\times {\mathcal D}^{m}, \end{aligned}$$

which gives the functional equation

$$\displaystyle \begin{aligned} D^{\leqslant m}(z)=1+z^2D^{\leqslant m-1}(z)D^{\leqslant m}(z). \end{aligned}$$

1.6

Let U denote a step $(1,2)$ and D a step $(1,-1)$. The first step must be U. By decomposing the path by the first passage through the line $y=1$ and then by the first passage through the line $y=0$, the path can be uniquely described as

$$\displaystyle \begin{aligned} {\mathcal A}=\epsilon +\{U\}\times {\mathcal A}\times \{D\}\times {\mathcal A}\times \{D\}\times {\mathcal A}. \end{aligned}$$

which gives the equation

$$\displaystyle \begin{aligned} A(z)=1+z^3A^3(z). \end{aligned}$$

The three paths of length six are depicted below.

1.7

If ${\mathcal N}$ denotes the class with a single node, then the class of trees such that the number of children at every node belongs to a subset $I\subset \mathbb N$ can be written as

$$\displaystyle \begin{aligned} {\mathcal T}_I={\mathcal N}\times (1+\sum_{i\in I}({\mathcal T}_I)^i). \end{aligned}$$

1.10

• i. There are $(n-3)$ diagonals and $n-2$ triangles.

• ii. Let ${\mathcal T}$ be the class of triangulations. Every triangulation can be described by a sequence of a triangulation, a triangle and a triangulation:

  $$\displaystyle \begin{aligned} {\mathcal T}=\{\epsilon\}+{\mathcal T}\times \{T\} \times {\mathcal T}. \end{aligned}$$

2.1

Hint: The class ${\mathcal C}$ of cycles has (exponential) generating function $C(z)=\log \frac {1}{1-z}$ and the class of even cycles is its even part

$$\displaystyle \begin{aligned} \frac{1}{2} (C(z)+C(-z))=\frac{1}{2}\left(\log\frac{1}{1-z}+\log \frac{1}{1+z}\right)=\log(\frac{1}{\sqrt{1-z^2}}). \end{aligned}$$

The class of permutations which have only even cycles in their cycle decomposition is ${\mathcal P}_{\text{even}}={\mathrm Set}({\mathcal C}_{\text{even}})$, ans its generating function is

$$\displaystyle \begin{aligned} P_{\text{even}}(z)=\exp(C_{\text{even}}(z))=\frac{1}{\sqrt{1-z^2}}=\sum_{n\geqslant 0} {-1/2\choose n}z^{2n}. \end{aligned}$$

2.5

The class ${\mathcal P}^{\text{even}}$ of permutations which decompose into an even number of cycles can be extracted from the construction $({\mathcal C})$ by selecting only the even powers:

$$\displaystyle \begin{aligned} {\mathcal P}^{\text{even}}=\sum_{k\geqslant 1} \frac{{\mathcal C}^{*2k}}{(2k)!} \end{aligned}$$

and the generating function is

$$\displaystyle \begin{aligned} P^{\text{even}}(z)=\cosh \left(\log \frac{1}{1-z}\right)=\frac{1}{2}\left(\frac{1}{1-z}+(1-z)\right). \end{aligned}$$

Its number is, for $n>1$,

$$\displaystyle \begin{aligned} n![z^n]P^{\text{even}}(z)=n!/2, \end{aligned}$$

as one may expect. 2.6

Hint: A word in ${\mathcal W}^{(k, r)}$ can be identified by the 4-tuple $(f^{-1}(a), f^{-1}(b), f^{-1}(\alpha ),$ $f^{-1}(\beta ))$ where $f:[n]\to \{a,b,\alpha ,\beta \}$ gives the positions of the letters in a word of length n, with the condition that $f^{-1}(a)$, $f^{-1}(b)$ have size at most k and $f^{-1}(\alpha )$ and $f^{-1}(\beta )$ have size at least r.

2.9

Hint: A labelled rooted star tree is a root together with a set of nodes which are the leaves. The class of rooted star trees can be described as ${\mathcal Z}*\mathsf { Set}({\mathcal Z})$. A forest of rooted labelled stars is

$$\displaystyle \begin{aligned} {\mathcal S}=\mathsf{Set}({\mathcal Z}*\mathsf{Set}({\mathcal Z})). \end{aligned}$$

2.10

• i. Let ${\mathcal U}_{2}$ be the class of urns of cardinality at least 2. We have

  $$\displaystyle \begin{aligned} {\mathcal P}_{k,2}=\overbrace{{\mathcal U}_{2}*\cdots *{\mathcal U}_{2}}^k/\sim_k. \end{aligned}$$

  where $\sim _k$ denotes the equivalence relation identifying two k-tuples that differ only in a permutation of its entries.

• ii. Every doubly surjective map $f:[n]\to [k]$ is identified by a sequence of k sets of cardinality at least two:

  $$\displaystyle \begin{aligned} {\mathcal S}_k=\overbrace{{\mathcal U}_2*\cdots*{\mathcal U}_2}^k. \end{aligned}$$

• iii. Every word of length n on the alphabet $\{a_1,\ldots ,a_k\}$ such that each symbol appears at least twice is identified with a doubly surjective map $f:[n]\to \{a_1,\ldots ,a_k\}$.

3.1

• i. $\frac {1}{24}(r^8+17r^4+6r^2)$.

• ii. $\frac {1}{24}(r^8+17r^4+6r^2)$.

3.2

• ii. $\frac {1}{12}(r^6+3r^4+4r^3+2r^2+2r).$

• iii. 74.

3.4

34.

3.6

• i. 700.

• ii. 414.

3.7

218.

3.9

• iii. 396.

• iv. No.

3.10

• iii. 104.

3.12

• i. 36.

• ii. The coefficient of $t^j$ in $15+11t+6t^2+3t^3+t^4$.

• iii. The coefficient of $t^j$ in

  $$\displaystyle \begin{aligned} \tfrac{1}{12}((1+t+t^3+t^5)^4+3(1+t^2+t^6+t^{10})^2+8(1+t^3+t^9+t^{15})(1+t+t^3+t^5)). \end{aligned}$$

3.13

• i. The number of 4-colourings is

  $$\displaystyle \begin{aligned} Z(4,4,4,4,4,4) \end{aligned}$$

  where

  $$\displaystyle \begin{aligned} Z(X_1,\ldots,X_6)=\frac{1}{12}(X_1^6+4X_2^3+2X_3^2+2X_6+3X_2^2X_1^2). \end{aligned}$$

• ii. The coefficient of $t^j$ in

  $$\displaystyle \begin{aligned} Z(t+3,t^2+3,t^3+3,t^4+3,t^5+3,t^6+3).\end{aligned}$$

• iii. The coefficient of $t^j$ in

  $$\displaystyle \begin{aligned} Z(t+2t^2+1,t^2+2t^4+1,t^3+2t^6+1,t^4+2t^8+1,t^5+2t^{10}+1,t^6+2t^{12}+1). \end{aligned}$$

4.1

• i. Prove that they are either ${\mathbb Z}/4{\mathbb Z}$ with addition and the second is ${\mathbb Z}/5{\mathbb Z}$ with multiplication.

• ii.

4.2

• iii. Let $q=2m$ and consider the latin square $\mathrm {L}_a$ defined on the elements of ${\mathbb F}_q$ with binary operation $x \circ y=ax+y$, where a is a non-zero element of ${\mathbb F}_q$.

  Recall that, by construction, ${\mathbb F}_q={\mathbb F}_2 [X] /(f)$, where f is an irreducible polynomial over ${\mathbb F}_2$ of degree $k+1$.

  Order the elements of ${\mathbb F}_q$ so that the first m are represented by polynomials of degree at most $k-1$. The top-left $m \times m$ array in $\mathrm {L}_1$ will contain only these elements, since summing two polynomials of degree at most $k-1$ gives a polynomial of degree at most $k-1$. By Theorem 4.19, $\mathrm {L}_1$ belongs to a set of $2m-1$ mutually orthogonal latin squares of order 2m.

4.9

Count triples $(x,y,\ell )$ where x and y are points incident with the line $\ell$ in two ways and apply the Erdős-De Bruijn theorem.

4.12

• i. $\{0,1,3\}$ is a difference set of ${\mathbb Z}/7{\mathbb Z}$ and $\{0,1,3,9 \}$ is a difference set of ${\mathbb Z}/13{\mathbb Z}$.

4.13

• i. $D=\{0,1,3\}$ is a relative difference set of ${\mathbb Z}/8{\mathbb Z}$.

5.1

Hint: Removing the leaves from the tree and their fathers, we get a smaller tree.

5.3

Hint: Let $U=A' \cup B'$ be a vertex cover, of minimum size where $A' \subseteq A$ and $B' \subseteq B$. Since U is a vertex cover, there are no edges between $A \setminus A'$ and $B \setminus B'$. Show that Hall’s condition fails for $A \setminus A'$.

5.4

Hint: Consider an Eulerian circuit C, substituting each vertex v by two vertices $v^-$ and $v^+$ and every edge $v_iv_{i+1}$ of C by the edge $v_i^+v_{i+1}^-$. The resulting graph is bipartite k-regular.

5.7

Consider the set of preferences in which, for $i \neq n$

$$\displaystyle \begin{aligned} x_i \ | \ y_i>y_{i-1}>\cdots y_1>y_{n-1}>\cdots y_{i+1}>y_n \end{aligned}$$

and for $x_n$

$$\displaystyle \begin{aligned} x_n \ | \ y_{n-1}>y_{n-2}>\cdots>y_1>y_n \end{aligned}$$

and for $i \neq n$,

$$\displaystyle \begin{aligned} y_i \ | \ x_{i-1}>x_{i-2}>\cdots x_1>x_n>x_{n-1}>\cdots > x_i. \end{aligned}$$

5.8

A 1-factorisation of the Petersen graph is equivalent to a proper 3-colouring of the edges. Up to change of colour there is a unique way to colour the outer edges, which leaves no choice for the colours of the outer to inner edges. But then the inner edges cannot be properly 3-coloured.

5.10

Let $X \cup A$ be the vertex partition of a bipartite graph $\Gamma$, where $|X|=|A|$.

Suppose that $\Gamma$ does not have a perfect matching. Then we will find a subset J of the vertices of A with the property that $N(J)$, the union of the neighbours of J, is smaller than J. This suffices to prove that if $|N(J)| \geqslant |J|$ for all subsets J of A then $\Gamma$ has a perfect matching.

Adding all edges between any two vertices of X, does not alter the property that $\Gamma$ does not have a perfect matching, since a perfect matching is a set of edges which have one end-vertex in A and the other in X. Hence, we can assume that the vertices X in $\Gamma$ form a complete subgraph.

Theorem 5.11 implies that there is a subset S of the vertices such that the number of odd components of $\Gamma \setminus S$ is larger than $|S|$. An odd component of $\Gamma \setminus S$ is either an isolated vertex in $N(S \cap X)$ or one other large component with vertex set C. Note that Hall’s condition implies that every vertex in A is connected to some vertex in X, so if there is a vertex in A not in C, then it is in $N(S \cap X)$.

Let J be the subset of $N(S \cap X)$ which are not in C. Then $N(J) \subseteq S \cap X$. Let $T=N(J)$.

Any vertex in $(S \cap X) \setminus T$ is joined to a vertex in C and so is in C. Its removal may change the status of C as an odd component, but it does not change the status of any other odd components. Therefore, it is possible that $\mathrm {oc}(\Gamma \setminus T)$ is one less than $\mathrm {oc}(\Gamma \setminus S)$ but then $|T|$ is one less than $|S|$, so we have that T also has the property that the number of odd components of $\Gamma \setminus T$ is more than $|T|$.

If $|C|$ is even then

$$\displaystyle \begin{aligned} |J|=\mathrm{oc}(\Gamma \setminus T) > |T|=|N(J)|. \end{aligned}$$

If $|C|$ is odd then $|T| \neq |J|$ (since $C=(A \cup X) \setminus (J \cup T)$ and $|A|=|X|$), so

$$\displaystyle \begin{aligned} |J|=\mathrm{oc}(\Gamma \setminus T) -1 \geqslant |T|=|N(J)|, \end{aligned}$$

implies $|J|>|N(J)|$.

5.11

• i. Counting vertices modulo 2.

• ii. By Tutte’s theorem, there exists a subset S of the vertices such that

  $$\displaystyle \begin{aligned} |\mathrm{oc}(\Gamma \setminus S)|-|S|>0. \end{aligned}$$

  By counting edge-vertex pairs, it follows that a cubic graph has an even number of vertices. Now apply i.

• iii. Suppose that $\Gamma$ does not have a perfect matching.

  By ii., there exists an S such that

  $$\displaystyle \begin{aligned} |\mathrm{oc}(\Gamma \setminus S) |-|S| \geqslant 2. \end{aligned}$$

  Each odd component of $\Gamma \setminus S$ has an odd number of vertices and so there are an odd number of edges from $\Gamma \setminus S$ to S. If this number is 1 then the edge is a bridge. If there are at most two bridges we reach a contradiction.

5.12

Hint: Consider what happens to the number of components with an odd number of edges if we remove an edge of a graph.

5.13

• i. In every unbalanced even component there are at least two vertices joined by the matching to a vertex in S, giving one direction. The converse follows by Tutte’s theorem.

• ii. If $\Gamma$ is bipartite then odd connected components must be unbalanced. Therefore,

  $$\displaystyle \begin{aligned} \mathrm{uc}(\Gamma)=\mathrm{oc}(\Gamma)+\mathrm{euc}(\Gamma)\leqslant \mathrm{oc}(\Gamma)+2\mathrm{euc}(\Gamma) \end{aligned}$$

  and the condition is necessary by part (i). Sufficiency again holds by Tutte’s theorem.

• iii. The graph $\Gamma =K_{1,2,3}$ has a perfect matching. Let S be one of the vertices in the stable set with two vertices. Then $\Gamma \setminus S$ has one odd component and one severely unbalanced component.

6.2

Hint: If $\Gamma / e$ is not 2-connected then the vertices of e have a neighbour in all components of $\Gamma \setminus e$, since $\Gamma$ is 2-connected.

Hint: If $\Gamma \setminus e$ is not 2-connected then it has a cut-vertex v. Since $\Gamma$ is 2-connected, the edge e must join vertices in distinct components of $\Gamma \setminus v$.

6.3

Hint: Suppose z is a cut-vertex for $\Gamma \setminus \{x,y\}$. Prove that $\{z, v_{xy}\}$ is a cut-set of $\Gamma /xy$, where $v_{xy}$ is the vertex obtained by contracting the edge xy.

Hint: Suppose $\{z,w\}$ is a cut-set of $\Gamma /xy$. Argue that $w=v_{xy}$.

6.4

Hint: By induction on k and apply Menger’s theorem.

6.6

Hint: Let $C_i$ be a connected component of $G \setminus S$ such that $C_i \cap S' = \emptyset$. Every vertex of S has a neighbour in $C_i$, since otherwise S is not a minimal separating set. Thus, $S \setminus S'$ is in the same connected component as $C_i$ in $G \setminus S'$.

6.9

Hint: We need to prove that a separator set has size at least n, where $|A|=n$ and A, B is the bipartite partition of the bipartite graph $\Gamma$. Suppose there is a separator set $S_A \cup S_B$ of size at most $n-1$, where $S_A \subseteq A$ and $S_B \subseteq B$. Prove that Hall’s condition fails for $J=A \setminus S_A$.

6.10

By induction on k. Note that $Q_k$ splits into two copies of $Q_{k-1}$, those whose first coordinate is zero and those whose first coordinate is 1. Argue that to separate $Q_k$ we must separate each of these copies of $Q_{k-1}$ and use that fact that $(1,x)$ is joined to $(0,x)$.

6.12

• iii. $K_3+E_5$.

7.1

The average degree is smaller than 6. If G is triangle free, the average degree is smaller than 4.

7.2

Every face is bounded by a cycle. If a cycle is not a face, consider the subgraph which has this cycle as outer face.

7.4

All vertices must be in the cycle of the outer face. Some vertex has degree two.

7.5

Add one vertex in the outer face connected to all vertices of $\Gamma$ and apply Kuratowski’s theorem.

7.6

• i. Use Euler’s formula and double counting faces and edges.

• ii. If the end vertices of an edge share more than two neighbours then $\Gamma$ has a separating triangle. Use induction on $|V(\Gamma )|$.

• iii. Suppose that $S=\{x,y\}$ separates $\Gamma$ and let X and Y  be two connected components of $\Gamma -S$. Contract edges until we eventually reach a maximal planar graph with four vertices and get a contradiction.

7.7

Let $H\subset \Gamma$ be a subdivision of $K_5$. If there is a vertex v not in H, there are three independent paths from v to three branching vertices of H and a subdivision of $K_{3,3}$ appears. Otherwise, choose a vertex $v\in V(H)$ in a path P joining two branching vertices x and y of H. There is a path from v to a third branching vertex different from x and y not using edges from P (by 3-connectedness) and again a subdivision of $K_{3,3}$ can be formed.

7.8

Let $S=\{x,y\}$ be a set of two vertices and $a, b$ arbitrary vertices from the skeleton different from x and y. By an appropriate projective transformation we may assume that the supporting planes through a and b are parallel. Let P be a plane parallel to them containing x and let z be a point on P interior to the polyhedra. The projection of the polyhedra to a plane orthogonal to the line xz sends $a,b$ and y to the boundary polygon of the projection and x to its interior. Since a polygon is 2-connected, there is a path joining the images of a and b avoiding the images of x and y. Lifting this path back to the polyhedra we obtain a path from a to b avoiding S.

7.9

First show that a maximal planar graph with an extra edge contains a subdivision of $K_5$. If xy is an edge not in G, consider the neighborhood of x in G. It induces a cycle (with x a wheel) with $k\geqslant 3$ vertices. Select three vertices in the boundary and three independent paths joining them to y (by 3-connectivity of G). This already gives $K_5^-$ (the complete graph minus one edge) as a subdivision in G. The added edge xy provides the subdivision of $K_5$. (It was proved by Mader that any graph with $3n-5$ edges contains $K_5$ as a minor, this is more difficult without the planarity condition of the graph minus one edge).

7.10

By induction on the number n of vertices, starting with $n=4$. Choose an edge $e=xy$ such that $\Gamma /e$ is maximal planar (a former exercise shows that such an edge exists). In a decomposition of three trees of $\Gamma /e$ consider the cases where the edges $av_{xy}$ and $v_{xy}b$, where a and b are the only common neighbours of x and y in $\Gamma$ and $v_{xy}$ is the vertex arising from the contraction, belong to the same tree or to two distinct trees. Show that the decomposition can be extended to G in both cases.

8.1

Order the vertices with a lex order from a given r-colouring of $\Gamma$. For the second part use an ordering where vertices joined by the edges of the matching are consecutive.

8.3

Colour $\chi (x,y)=\chi _{\Gamma }(x)+\chi _{\Gamma '}(y) \pmod {\max (\chi (\Gamma ), \chi (\Gamma ')\}}$.

8.4

There is a graph homomorphism from $\Gamma \times \Gamma '$ to $K_r$ with $r=\min \{\chi (\Gamma ), \chi (\Gamma ')\}$.

8.5

An $(r-1)$ colouring could be extended from the connected components of an $(r-2)$ edge-cut.

8.8

Given an r-colouring orient every edge from smaller to larger colour of its end vertices. For the converse, given the orientation, show that colouring v with the length of the longest oriented path ending at v provides a proper colouring.

8.9

Hint: Use the first part of the exercise and induction.

8.10

[iii.] $f_{K_n}(x)=\prod _{i=0}^{n-1} (x-i)$, $f_T(x)=x(x-1)^{n-1}$.

8.11

If the color classes are $C_1,\ldots ,C_r$ then $m\leqslant \prod _{i<j}|C_i|\cdot |C_j|$. This function is maximized under $\sum _i|C_i|=n$ by taking all $C_i$ with almost the same size.

8.12

It is 2-degenerated.

8.13

Every colour class induces a matching and misses at least one vertex.

8.14

Each connected component of $\Gamma -e$ with e a bridge has odd order.

8.15

For n odd label the vertices $\{0,1,\ldots n-1\}$ and colour the edges $\chi (xy)=x+y \pmod {n}$. For n even colour $K_{n-1}$ and add one vertex adjacent to all of them. For each vertex of $K_{n}$ one colour is available for the edge joining it to the new vertex.

8.17

For $k\geqslant 2$ let $n={2k-1\choose k}$ and consider the list assignment of $K_{n,n}$ where vertices in each stable set have pairwise distinct k-subsets of $\{1,\ldots ,n\}$. There is no list colouring with such an assignment of lists.

8.20

• i. Run the greedy coloring algorithm on the following order: $x_n$ is a vertex with degree at most k; given $x_{n},\ldots ,x_{n-i+1}$ choose a vertex of degree of degree at most k in $G[V\setminus \{ x_n,\ldots ,x_{n-i+1}\}]$ (which exists by degeneracy). At each step of the algorithm at most k colors are forbidden.

• ii. Prove by induction. Note that we can add an edge from x to $G-x$ to obtain a k-degenerated graph $G'$ (every subgraph containing x has $\delta (G')\leqslant d_{G'}(x)=k$ while a subgraph not containing x is still k degenerated).

• iii. An outer planar graph is 2-degenerate.

9.4

Hint: Consider subsets of size $n_0$ of the vertices of $\Gamma$. Prove that at least $\frac {1}{2}\epsilon {n \choose n_0}$ of them have at least $(\rho +\frac {1}{2}\epsilon ){n_0 \choose 2}$ edges.

9.5

Each vertex has $m+1$ neighbours unless x is an absolute point, i.e. $b(u,u)=0$, in which case $x=\langle u \rangle$ has m neighbours. There are $m+1$ absolute points.

9.6

Consider two vertices of the graph u and w. The common neighbours of u and w are points which are incident with the planes $\sigma (u)$ and $\sigma (w)$. These two planes intersect in a line. A line intersects the ellipsoidal cone either in at most two points or lies on the cone. Note that the planes $\sigma (u)$ and $\sigma (w)$ are not incident with the origin, so the line is not lying on the cone. Therefore, u and w have at most two common neighbours and so G contains no $K_{2,3}$ subgraph.

9.7

Let $\Gamma$ be a bipartite graph on n vertices which contains no $K_{t,s}$. Let $V_0$ be the least large of the two stable sets, so $n_0=|V_0| \leqslant \frac {1}{2}n$. Let N be the number of copies of $K_{1,t}$ in which t of the vertices are in $V_0$.

Then, since $\Gamma$ contains no $K_{t,s}$,

$$\displaystyle \begin{aligned} N \leqslant {n_0 \choose t}(s-1). \end{aligned}$$

Furthermore,

$$\displaystyle \begin{aligned} N=\sum_{u \in V_0} {d(u) \choose t} \geqslant n_0 { \lfloor (m/n_0) \rfloor t} \geqslant n_0 ((m/n_0)-t)^t/t!. \end{aligned}$$

Combine these two inequalities and take $n_0$ large enough,

9.8

Hint: consider AG$(3,q)$ as a linear space.

9.10

Construct a graph whose vertices are the points of S and where two vertices are joined by an edge if and only if the distance between them in in D.