\magnification 1400 \parskip 10pt \parindent 0pt \hoffset -0.3truein \hsize 7truein \def\N{{\bf N}} \def\S{\Sigma_{i=1}^n} \def\Sm{\Sigma_{i=1}^m} \def\R{{\bf R}} \def\C{{\bf C}} \def\x{{\bf x}} \def\a{{\bf a}} \def\y{{\bf y}} \def\e{{\bf e}} \def\i{{\bf i}} \def\j{{\bf j}} \def\k{{\bf k}} \def\l{{\it l}} \def\ep{\epsilon} \def\v{\vskip 50pt} \def\u{\vskip -5pt} \def\h{\hskip 10pt} \def\hb{\hfil \break} \def\hr{\vskip 5pt \hrule \vskip -5pt} Let $N(d_1, ..., d_n) = $ the number of labeled trees with $n$ vertices $\{v_1, ..., v_n\}$ such that $deg(v_i) = d_i + 1$. Let $C(n-2, d_1, ..., d_n) = {(n-2)! \over d_1! d_2! \cdot\cdot\cdot d_n!}$. section 3.5 {\bf 34e.)} Claim: $N(d_1, ..., d_n) = \cases{ C(n-2; d_1, ..., d_n) & if $\Sigma d_i = n-2$ \cr 0 & otherwise}$ Claim $N(d_1, ..., d_n) = 0$ if $\Sigma d_i \not= n-2$ Proof: \hb $\Sigma d_i = \Sigma_{i=1}^{n} (deg(v_i) - 1) = [\Sigma_{i=1}^{n} (deg(v_i)] - n = [2(n-1)] - n = 2n- 2 - n = n - 2$ Thus $\Sigma d_i = n-2$. Hence $N(d_1, ..., d_n) = 0$ if $\Sigma d_i \not= n-2$. Claim $N(d_1, ..., d_n) = C(n-2; d_1, ..., d_n) $ if $\Sigma d_i = n-2$ ~~~~(*) Proof by induction on $k$ = number of vertices. By part $a$, the equality holds for $n = 2$. Induction hypothesis: Suppose (*) is true when $k = n-1$. By part b, $d_j =0$ for some $j$. WLOG assume $j = n$. Thus by part c, $N(d_1, ..., d_n) = N(d_1, ..., d_{n-1}, 0) = N(d_1 - 1, d_2, ..., d_{n-1}) + N(d_1, d_2 - 1, d_3..., d_{n-1}) + ... + N(d_1, ...,d_{n-2}, d_{n-1} - 1)$. By part d, $C(n-2; d_1, ..., d_n) = C(n-2; d_1, ..., d_{n-1}, 0) = C(n-3; d_1 - 1, d_2, ..., d_{n-1}) + C(n-3; d_1, d_2 - 1, d_3..., d_{n-1}) + ... + C(n-3; d_1, ...,d_{n-2}, d_{n-1} - 1)$. By the induction hypothesis, $N(d_1, ..., d_n) = N(d_1, ..., d_{n-1}, 0) = N(d_1 - 1, d_2, ..., d_{n-1}) + N(d_1, d_2 - 1, d_3..., d_{n-1}) + ... + N(d_1, ...,d_{n-2}, d_{n-1} - 1) = $ $C(n-3; d_1 - 1, d_2, ..., d_{n-1}) + C(n-3; d_1, d_2 - 1, d_3..., d_{n-1}) + ... + C(n-3; d_1, ...,d_{n-2}, d_{n-1} - 1)= C(n-2; d_1, ..., d_{n-1}, 0) = C(n-2; d_1, ..., d_n)$. {\bf 34a).} Suppose $n = 2$. A tree with 2 vertices has 1 edge. Thus $deg(v_i) = 1$ for $i = 1, 2$. Thus $d_i = 0$ for $i = 1, 2$. There is exactly one labeled tree with 2 vertices, $T = ( \{v_1, v_2\}, \{ \{v_1, v_2\} \} )$. Thus $N(0, 0) = 1$. ~~ $C(0; 0, 0) = {0! \over 0! 0! } = 1$. ~~Thus (*) holds for $n = 2$. {\bf 34b.)} Claim $d_i = 0$ for some $i$. Suppose $d_i > 0$ for all $i$. Then $deg(v_i) = d_i + 1 > 1$ for all $i$. That is, $deg(v_i) \geq 2$ for all $i$. The number of edges in a graph = ${1 \over 2} \Sigma deg(v_i)$. The number of edges in a tree with $n$ vertices is $n - 1$. Thus $n-1 = {1 \over 2} \Sigma_{i = 1}^n deg(v_i) \geq {1 \over 2} \Sigma_{i = 1}^n 2 = {1 \over 2} (2n) = n$, a contradiction. Thus $d_i = 0$ for some $i$. {\bf 34c).} Let $A$ = set of all labeled trees with $n$ vertices $\{v_1, ..., v_n\}$ such that $deg(v_i) = d_i + 1$. Then $|A| = N(d_1, ..., d_n)$. For $j = 1, ..., n-1$, let $B_j$ = set of all labeled trees with $n -1$ vertices $\{v_1, ..., v_{n-1} \}$ such that $deg(v_i) = d_i + 1$, $i \not= j$ and $deg(v_j) = (d_j - 1) + 1$. Then $|B_j| = N(d_1, ..., d_{j-1}, d_j - 1, d_{j+1}, ..., d_{n-1}) $ for $j = 1, ..., n-1$. Note if $T_j \in B_j$, then $deg(v_j) = d_j$. Suppose $k \not= j$. If $T_k \in B_k$, then $deg(v_j) = d_j + 1$. Thus $T_j$ is not isomorphic to $T_j$. Thus $B_j \cap B_k = \emptyset$ for $k \not= j$. Claim: There exists a bijection $f: A \rightarrow \cup_{i=1}^{n-1} B_i$. Note if the claim is true, then $|A| = |\cup_{i=1}^{n-1} B_i| = \Sigma_{i=1}^{n-1}|B_i|$, since the $B_i $ are pairwise disjoint. Define $g: \cup_{i=1}^{n-1} B_i \rightarrow A$. Let $T = (V, E)\in B_j$. Let $g(T) = (V \cup \{v_n\}, E \cup \{ \{v_j, v_n \} \})$. Note $g(T)$ has $n$ vertices and $deg(v_i) = d_i + 1$ for $i = 1, ..., n$. Thus $g: \cup_{i=1}^{n-1} B_i \rightarrow A$ is well-defined. Claim $g^{-1}$ exists. WLOG assume $d_n = 0$ (relabel the vertices if needed). $d_n = 0$ implies $deg(v_n) = 1$. Suppose the vertex adjacent to $v_n$ is labeled $v_j$. Let $T' (V', E') \in A$. Define $f: A \rightarrow \cup_{i=1}^{n-1} B_i$ by $f(T') = (V'- \{v_n\}, E' - \{ \{v_j, v_n \} \})$. Note that $f(T')$ has $n-1$ vertices, $\{v_1, ..., v_{n-1}\}$ and $deg(v_j) = d_j$, $deg(v_i) = d_i+1$ for $i \not= j$. Thus $f(T')$ is in $B_j$, and hence $f$ is well-defined. $f(g((V, E))) = f( (V \cup \{v_n\}, E \cup \{ \{v_j, v_n \} \}) = (V, E)$. $g(f((V, E))) = g( (V - \{v_n\}, E - \{ \{v_j, v_n \} \} ) = (V, E)$. Thus $g$ is invertible. Thus $g$ is a bijection. Thus $|A| = |\cup_{i=1}^{n-1} B_i| = \Sigma_{i=1}^{n-1}|B_i|$. Alternate proof that $g$ is a bijection: Claim: $g$ is onto. Let $T' = (V', E') \in A$. Let $T = (V'- \{v_n\}, E' - \{ \{v_j, v_n \} \})$. Then $g(T) = g( (V' - \{v_n\}, E' - \{ \{v_j, v_n \} \} ) = (V', E') = T'$. Claim $g$ is 1-1: Suppose $g(T) = g(S)$. Claim $T$ and $S$ are isomorphic ... \vfill \vfill \vfill \vfill \vfill \vfill \vskip 20pt {\bf 34d).} Note that by the right-hand side of the equation, we are given that $\Sigma_{i=1}^{n-1} d_i = [\Sigma_{i=1}^{n} d_i] - d_n = n- 2 - 0 = n-2$ $\Sigma_{i=1}^{n-1} C(n-3; d_1, ..., d_{i-1}, d_i - 1, d_{i+1}, ..., d_{n-1}) = \Sigma_{i=1}^{n-1} { (n-3)! \over d_1! \cdot \cdot \cdot d_{i-1}!, (d_i - 1)!, d_{i+1}! \cdot \cdot \cdot d_{n-1}! }$ $= \Sigma_{i=1}^{n-1} { (n-3)! d_i\over d_1! \cdot \cdot \cdot d_{i-1}!, d_i!, d_{i+1}! \cdot \cdot \cdot d_{n-1}! }$ $= { (n-3)! \over d_1! \cdot \cdot \cdot d_{i-1}!, d_i!, d_{i+1}! \cdot \cdot \cdot d_{n-1}! } \Sigma_{i=1}^{n-1} d_i$ $= { (n-3)! \over d_1! \cdot \cdot \cdot d_{n-1}! } (n-2)$ $= { (n-2)! \over d_1! \cdot \cdot \cdot d_{n-1}! 0!}$ $= { (n-2)! \over d_1! \cdot \cdot \cdot d_{n-1}! d_n!}$ $= C(n-2, d_1, ..., d_n)$ \end