As noted above, we convert this problem into an equivalent graph theory problem. Let $[n]$ denote the set of the first $n$ positive integers; i.e. $[n]$ denotes $\{1,2,\cdots,n\}$. Let $V$ denote the set $[n] \times \{\text{red},\text{blue},\text{green}\}$ (here the Cartesian product of sets is meant). For any element $v = (i,c) \in V$, we will define the rank of $v$ to be $i$, and the color of $v$ to be $c$. (For example, the rank of $(3,\text{red})$ is $3$, and the color of $(7,\text{green})$ is $\text{green}$.) Suppose $G$ is a simple undirected graph with vertex set $V$. We say that $G$ is induced-Eulerian if no two vertices of the same rank are adjacent in $G$, and if for any $i,j \in [n]$ with $i \not= j$, the induced subgraph $G_{ij}$ on $\{i,j\} \times \{\text{red},\text{blue},\text{green}\}$ has all vertices of even degree. Define a color-assignment to be a function $f:[n] \rightarrow \{\text{red},\text{blue},\text{green}\}$. If $G$ is induced-Eulerian, a color-assignment $f$ is a Eulerian-assignment on $G$ if the induced subgraph $H(f)$ on $\{(i,f(i)):i \in [n]\}$ has all vertices of even degree. [asy][asy] unitsize(3cm); pen tp = linewidth(2); real ep = 12.0; real r = 0.05; real t0 = 0.0; real t1 = 60.0; real t2 = 120.0; real t3 = 180.0; real t4 = 240.0; real t5 = 300.0; pair a0 = (Cos(t0-ep),Sin(t0-ep)); pair b0 = (Cos(t0),Sin(t0)); pair c0 = (Cos(t0+ep),Sin(t0+ep)); pair a1 = (Cos(t1-ep),Sin(t1-ep)); pair b1 = (Cos(t1),Sin(t1)); pair c1 = (Cos(t1+ep),Sin(t1+ep)); pair a2 = (Cos(t2-ep),Sin(t2-ep)); pair b2 = (Cos(t2),Sin(t2)); pair c2 = (Cos(t2+ep),Sin(t2+ep)); pair a3 = (Cos(t3-ep),Sin(t3-ep)); pair b3 = (Cos(t3),Sin(t3)); pair c3 = (Cos(t3+ep),Sin(t3+ep)); pair a4 = (Cos(t4-ep),Sin(t4-ep)); pair b4 = (Cos(t4),Sin(t4)); pair c4 = (Cos(t4+ep),Sin(t4+ep)); pair a5 = (Cos(t5-ep),Sin(t5-ep)); pair b5 = (Cos(t5),Sin(t5)); pair c5 = (Cos(t5+ep),Sin(t5+ep)); dot(a0); dot(b0); dot(c0); dot(a1); dot(b1); dot(c1); dot(a2); dot(b2); dot(c2); dot(a3); dot(b3); dot(c3); dot(a4); dot(b4); dot(c4); dot(a5); dot(b5); dot(c5); draw(circle(b0,r)); draw(circle(c1,r)); draw(circle(a2,r)); draw(circle(c3,r)); draw(circle(c4,r)); draw(circle(a5,r)); draw(b0--c1, p = tp); draw(b0--c3, p = tp); draw(b0--c4, p = tp); draw(b0--a5, p = tp); draw(c1--a2, p = tp); draw(a2--c4, p = tp); draw(c3--a5, p = tp); draw(b0--c1--a0--a1--cycle); draw(a0--a2--c0--b2--cycle); draw(b0--c3--c0--a3--a0--b3--cycle); draw(b0--c4--a0--b4--cycle); draw(b0--a5--a0--c5--cycle); draw(c1--a2--b1--c2--cycle); draw(a1--a3--c1--b3--b1--c3--cycle); draw(b1--c4--a1--b4--cycle); draw(c1--c5--b1--b5--cycle); draw(a2--b3--c2--a3--cycle); draw(a2--c4--b2--b4--c2--a4--cycle); draw(a2--b5--b2--a5--c2--c5--cycle); draw(c3--a5--a3--c5--cycle); [/asy][/asy] Figure 1. An example of an Eulerian-assignment on an induced-Eulerian graph when $n = 6$. Let $\text{ColAsn}$ denote the set of color-assignments (so it has cardinality $3^n$). We have the following Theorem: Theorem. If $G$ is induced-Eulerian, then it has an odd number of Eulerian-assignments. To show that a proof of Theorem would solve the original problem, label the $3n$ triangle-edges with distinct elements of $V$ such that in each triangle, the $3$ triangle-edges have the same rank. [asy][asy] unitsize(2cm); draw((Cos(330),Sin(330))--(0,1)--(Cos(210),Sin(210))--cycle); label("$(i,$ red$)$",(0.8,0.3)); label("$(i,$ blue$)$",(-0.8,0.3)); label("$(i,$ green$)$",(0,-0.6)); [/asy][/asy] Figure 2. For each triangle, choose an unused $i \in [n]$ and label its three triangle-edges with $(i,\text{red}),(i,\text{blue}),(i,\text{green})$ in some order. Construct a graph $G$ whose vertices are the $3n$ triangle-edges, and two vertices are adjacent in $G$ if and only if the corresponding triangle-edges intersect (it does not count if they intersect at an endpoint). [asy][asy] unitsize(2cm); pair rs = (2,0); pair ds = (0,-3); pair left = (-1,0); pair right = (1,0); pair up = (0,1); pair down = (0,-1); real th1 = 55.0; real th2 = 140.0; real eps = 0.2; draw(left+eps*(Cos(th1/2),Sin(th1/2))--left--right--right+eps*(Cos(90+th1/2),Sin(90+th1/2))); label("$(i,$ red$)$",(0.7,-0.15)); draw(up+eps*(Cos(135+th2/2),Sin(135+th2/2))--up--down--down+eps*(Cos(45+th2/2),Sin(45+th2/2))); label("$(j,$ blue$)$",(0.4,-0.8)); dot(rs+(0,0.5)); dot(rs+(0,-0.5)); draw(rs+(0,0.5)--rs+(0,-0.5)); label("$(i,$ red$)$",rs+(0,0.65)); label("$(j,$ blue$)$",rs+(0,-0.65)); draw(up+eps*(Cos(-45+th1/2),Sin(-45+th1/2))+ds--up+ds--right+ds--right+eps*(Cos(90+th1/2),Sin(90+th1/2))+ds); label("$(i,$ red$)$",(0.2,0.3)+ds); draw(left+eps*(Cos(th2/2),Sin(th2/2))+ds--left+ds--down+ds--down+eps*(Cos(45+th2/2),Sin(45+th2/2))+ds); label("$(j,$ blue$)$",(-0.2,-0.3)+ds); dot(rs+ds+(0,0.5)); dot(rs+ds+(0,-0.5)); //draw(rs+ds+(0,0.5)--rs+ds+(0,-0.5)); label("$(i,$ red$)$",rs+ds+(0,0.65)); label("$(j,$ blue$)$",rs+ds+(0,-0.65)); [/asy][/asy] Figure 3. The vertices $(i,\text{red})$ and $(j,\text{blue})$ will be adjacent in $G$ if and only if the two triangle-edges corresponding to $(i,\text{red})$ and $(j,\text{blue})$ intersect. Then $G$ is induced-Eulerian, since each induced subgraph $G_{ij}$ is either a $6$-cycle, a $4$-cycle, or an empty graph. [asy][asy] unitsize(2cm); real vsp = 0.3; draw((1,0)--(-0.5,Sin(120))--(-0.5,Sin(240))--cycle); draw((0.5,Sin(60))--(-1,0)--(0.5,Sin(300))--cycle); label("$a$",(0.8,0.25)); label("$b$",(-0.6,0.7)); label("$c$",(0.8,-0.25)); label("$x$",(-0.8,-0.25)); label("$y$",(0.6,0.7)); label("$z$",(-0.8,0.25)); dot((2,vsp)); label("$a$",(1.9,vsp)); dot((2,0)); label("$b$",(1.9,0)); dot((2,-vsp)); label("$c$",(1.9,-vsp)); dot((3,vsp)); label("$x$",(3.1,vsp)); dot((3,0)); label("$y$",(3.1,0)); dot((3,-vsp)); label("$z$",(3.1,-vsp)); draw((2,vsp)--(3,0)--(2,-vsp)--(3,vsp)--(2,0)--(3,-vsp)--cycle); [/asy][/asy] [asy][asy] unitsize(2cm); real vsp = 0.3; draw((1,0)--(-0.5,Sin(120))--(-1,0)--cycle); draw((0.5,Sin(60))--(-0.5,Sin(240))--(0.5,Sin(300))--cycle); label("$a$",(0.8,-0.1)); label("$b$",(0.8,0.25)); label("$c$",(-0.7,0.75)); label("$x$",(0.6,0.7)); label("$y$",(0.3,0.75)); label("$z$",(0.4,-0.95)); dot((2,vsp)); label("$a$",(1.9,vsp)); dot((2,0)); label("$b$",(1.9,0)); dot((2,-vsp)); label("$c$",(1.9,-vsp)); dot((3,vsp)); label("$x$",(3.1,vsp)); dot((3,0)); label("$y$",(3.1,0)); dot((3,-vsp)); label("$z$",(3.1,-vsp)); draw((2,vsp)--(3,vsp)--(2,0)--(3,0)--cycle); [/asy][/asy] [asy][asy] unitsize(2cm); real vsp = 0.3; draw((1,0)--(0.5,Sin(60))--(-0.5,Sin(120))--cycle); draw((-1,0)--(-0.5,Sin(240))--(0.5,Sin(300))--cycle); label("$a$",(1,0.2)); label("$b$",(-0.4,1)); label("$c$",(-0.4,0.7)); label("$x$",(-1,-0.2)); label("$y$",(0.4,-1)); label("$z$",(0.4,-0.7)); dot((2,vsp)); label("$a$",(1.9,vsp)); dot((2,0)); label("$b$",(1.9,0)); dot((2,-vsp)); label("$c$",(1.9,-vsp)); dot((3,vsp)); label("$x$",(3.1,vsp)); dot((3,0)); label("$y$",(3.1,0)); dot((3,-vsp)); label("$z$",(3.1,-vsp)); [/asy][/asy] Figure 4. No matter how two triangles intersect, they will correspond to induced subgraphs in which every vertex has even degree. Then Theorem tells us that there are an odd number of Eulerian-assignments on $G$; hence, there exists an Eulerian-assignment on $G$. Observe that Eulerian-assignments on $G$ correspond to marking exactly one triangle-edge from each triangle, such that each marked edge intersects an even number of other marked edges. Since there exists an Eulerian-assignment on $G$, there exists such a marking, and once we have such a marking, we finish by declaring that each boy and each girl will sit at an endpoint of some marked edge, and then placing the boys and girls in an alternating manner. The marking guarantees that for each marked edge, one of its endpoints will be occupied by a boy and the other endpoint will be occupied by a girl. Note that each triangle contains exactly one marked edge. Thus, for each triangle, one of its vertices will be occupied by a boy, one of its vertices will be occupied by a girl, and the remaining vertex will be vacant, as desired. [asy][asy] unitsize(2cm); pen tp = linewidth(2); pair b0 = (Cos(0),Sin(0)); pair g1 = (Cos(18),Sin(18)); pair b2 = (Cos(18*2),Sin(18*2)); pair g3 = (Cos(18*3),Sin(18*3)); pair b4 = (Cos(18*4),Sin(18*4)); pair g5 = (Cos(18*5),Sin(18*5)); pair b6 = (Cos(18*6),Sin(18*6)); pair g7 = (Cos(18*7),Sin(18*7)); pair b8 = (Cos(18*8),Sin(18*8)); pair g9 = (Cos(18*9),Sin(18*9)); pair b10 = (Cos(18*10),Sin(18*10)); pair g11 = (Cos(18*11),Sin(18*11)); pair b12 = (Cos(18*12),Sin(18*12)); pair g13 = (Cos(18*13),Sin(18*13)); pair b14 = (Cos(18*14),Sin(18*14)); pair g15 = (Cos(18*15),Sin(18*15)); pair b16 = (Cos(18*16),Sin(18*16)); pair g17 = (Cos(18*17),Sin(18*17)); pair b18 = (Cos(18*18),Sin(18*18)); pair g19 = (Cos(18*19),Sin(18*19)); draw(b0--g13, p = tp); draw(b2--g15, p = tp); draw(b4--g7, p = tp); draw(b6--g19, p = tp); draw(b8--g9, p = tp); draw(b10--g17, p = tp); draw(b12--g3, p = tp); draw(b14--g11, p = tp); draw(b16--g5, p = tp); draw(b18--g1, p = tp); label("B",1.1*b0); label("G",1.1*g1); label("B",1.1*b2); label("G",1.1*g3); label("B",1.1*b4); label("G",1.1*g5); label("B",1.1*b6); label("G",1.1*g7); label("B",1.1*b8); label("G",1.1*g9); label("B",1.1*b10); label("G",1.1*g11); label("B",1.1*b12); label("G",1.1*g13); label("B",1.1*b14); label("G",1.1*g15); label("B",1.1*b16); label("G",1.1*g17); label("B",1.1*b18); label("G",1.1*g19); pair rs = (2.5,0); pair p0 = (Cos(0),Sin(0)); pair p1 = (Cos(12),Sin(12)); pair p2 = (Cos(12*2),Sin(12*2)); pair p3 = (Cos(12*3),Sin(12*3)); pair p4 = (Cos(12*4),Sin(12*4)); pair p5 = (Cos(12*5),Sin(12*5)); pair p6 = (Cos(12*6),Sin(12*6)); pair p7 = (Cos(12*7),Sin(12*7)); pair p8 = (Cos(12*8),Sin(12*8)); pair p9 = (Cos(12*9),Sin(12*9)); pair p10 = (Cos(12*10),Sin(12*10)); pair p11 = (Cos(12*11),Sin(12*11)); pair p12 = (Cos(12*12),Sin(12*12)); pair p13 = (Cos(12*13),Sin(12*13)); pair p14 = (Cos(12*14),Sin(12*14)); pair p15 = (Cos(12*15),Sin(12*15)); pair p16 = (Cos(12*16),Sin(12*16)); pair p17 = (Cos(12*17),Sin(12*17)); pair p18 = (Cos(12*18),Sin(12*18)); pair p19 = (Cos(12*19),Sin(12*19)); pair p20 = (Cos(12*20),Sin(12*20)); pair p21 = (Cos(12*21),Sin(12*21)); pair p22 = (Cos(12*22),Sin(12*22)); pair p23 = (Cos(12*23),Sin(12*23)); pair p24 = (Cos(12*24),Sin(12*24)); pair p25 = (Cos(12*25),Sin(12*25)); pair p26 = (Cos(12*26),Sin(12*26)); pair p27 = (Cos(12*27),Sin(12*27)); pair p28 = (Cos(12*28),Sin(12*28)); pair p29 = (Cos(12*29),Sin(12*29)); pair tb0 = p0; pair tg0 = p18; pair tt0 = p5; pair tb1 = p2; pair tg1 = p21; pair tt1 = p23; pair tb2 = p4; pair tg2 = p9; pair tt2 = p8; pair tb3 = p7; pair tg3 = p29; pair tt3 = p19; pair tb4 = p10; pair tg4 = p12; pair tt4 = p16; pair tb5 = p14; pair tg5 = p26; pair tt5 = p22; pair tb6 = p17; pair tg6 = p3; pair tt6 = p11; pair tb7 = p20; pair tg7 = p15; pair tt7 = p25; pair tb8 = p24; pair tg8 = p6; pair tt8 = p13; pair tb9 = p28; pair tg9 = p1; pair tt9 = p27; draw(tb0+rs--tg0+rs, p = tp); draw(tg0+rs--tt0+rs--tb0+rs); label("B",1.1*tb0+rs); label("G",1.1*tg0+rs); draw(tb1+rs--tg1+rs, p = tp); draw(tg1+rs--tt1+rs--tb1+rs); label("B",1.1*tb1+rs); label("G",1.1*tg1+rs); draw(tb2+rs--tg2+rs, p = tp); draw(tg2+rs--tt2+rs--tb2+rs); label("B",1.1*tb2+rs); label("G",1.1*tg2+rs); draw(tb3+rs--tg3+rs, p = tp); draw(tg3+rs--tt3+rs--tb3+rs); label("B",1.1*tb3+rs); label("G",1.1*tg3+rs); draw(tb4+rs--tg4+rs, p = tp); draw(tg4+rs--tt4+rs--tb4+rs); label("B",1.1*tb4+rs); label("G",1.1*tg4+rs); draw(tb5+rs--tg5+rs, p = tp); draw(tg5+rs--tt5+rs--tb5+rs); label("B",1.1*tb5+rs); label("G",1.1*tg5+rs); draw(tb6+rs--tg6+rs, p = tp); draw(tg6+rs--tt6+rs--tb6+rs); label("B",1.1*tb6+rs); label("G",1.1*tg6+rs); draw(tb7+rs--tg7+rs, p = tp); draw(tg7+rs--tt7+rs--tb7+rs); label("B",1.1*tb7+rs); label("G",1.1*tg7+rs); draw(tb8+rs--tg8+rs, p = tp); draw(tg8+rs--tt8+rs--tb8+rs); label("B",1.1*tb8+rs); label("G",1.1*tg8+rs); draw(tb9+rs--tg9+rs, p = tp); draw(tg9+rs--tt9+rs--tb9+rs); label("B",1.1*tb9+rs); label("G",1.1*tg9+rs); pair ls = (-2.5,0); draw(tb0+ls--tg0+ls, p = tp); draw(tg0+ls--tt0+ls--tb0+ls); draw(tb1+ls--tg1+ls, p = tp); draw(tg1+ls--tt1+ls--tb1+ls); draw(tb2+ls--tg2+ls, p = tp); draw(tg2+ls--tt2+ls--tb2+ls); draw(tb3+ls--tg3+ls, p = tp); draw(tg3+ls--tt3+ls--tb3+ls); draw(tb4+ls--tg4+ls, p = tp); draw(tg4+ls--tt4+ls--tb4+ls); draw(tb5+ls--tg5+ls, p = tp); draw(tg5+ls--tt5+ls--tb5+ls); draw(tb6+ls--tg6+ls, p = tp); draw(tg6+ls--tt6+ls--tb6+ls); draw(tb7+ls--tg7+ls, p = tp); draw(tg7+ls--tt7+ls--tb7+ls); draw(tb8+ls--tg8+ls, p = tp); draw(tg8+ls--tt8+ls--tb8+ls); draw(tb9+ls--tg9+ls, p = tp); draw(tg9+ls--tt9+ls--tb9+ls); [/asy][/asy] Figure 5. Given a marking, we can place the boys and girls in an alternating manner such that for each marked edge, one of its endpoints will be occupied by a boy and the other endpoint will be occupied by a girl. Now we prove Theorem. Proof of Theorem. Let $G$ be fixed throughout. For each $i,j \in [n]$ with $i\not= j$, let $X_{ij}: \text{ColAsn} \rightarrow \mathbb{F}_2$ be the function defined in the following manner: $X_{ij}(f)$ equals $1$ if $(i,f(i))$ and $(j,f(j))$ are adjacent in $G$, and $X_{ij}(f)$ equals $0$ otherwise. [asy][asy] unitsize(2cm); pair rs = (3,0); real vsp = 0.3; real r = 0.05; dot((0,vsp)); dot((0,0)); dot((0,-vsp)); dot((1,vsp)); dot((1,0)); dot((1,-vsp)); draw(circle((0,0),r)); label("$(i,f(i))$",(-0.45,0)); draw(circle((1,-vsp),r)); label("$(j,f(j))$",(1.45,-vsp)); draw((0,vsp)--(1,0)--(0,-vsp)--(1,vsp)--(0,0)--(1,-vsp)--cycle); label("$X_{ij}(f) = 1$",(0.5,-vsp-0.3)); dot((0,vsp)+rs); dot((0,0)+rs); dot((0,-vsp)+rs); dot((1,vsp)+rs); dot((1,0)+rs); dot((1,-vsp)+rs); draw(circle((0,0)+rs,r)); label("$(i,f(i))$",(-0.45,0)+rs); draw(circle((1,0)+rs,r)); label("$(j,f(j))$",(1.45,0)+rs); draw((0,vsp)+rs--(1,0)+rs--(0,-vsp)+rs--(1,vsp)+rs--(0,0)+rs--(1,-vsp)+rs--cycle); label("$X_{ij}(f) = 0$",(0.5,-vsp-0.3)+rs); [/asy][/asy] Figure 6. A color-assignment $f \in \text{ColAsn}$ corresponds to a way to circle one vertex from each rank. We have $X_{ij}(f) = 1$ if and only if the circled vertices from ranks $i$ and $j$ are adjacent in $G$. (The graph depicted on each side of the Figure is the induced subgraph $G_{ij}$.) Now define the function $s: \text{ColAsn} \rightarrow \mathbb{F}_2$ as follows: \[s(f) = \prod_i \left(1+\sum_{j \not= i}X_{ij}(f)\right) = (1+X_{12}(f)+X_{13}(f)+X_{14}(f)+\cdots)(X_{21}(f)+1+X_{23}(f)+X_{24}(f)+\cdots)\cdots\]for all $f \in \text{ColAsn}$. Observe that $s(f) = 1$ if and only if $f$ is an Eulerian-assignment on $G$. Thus, it suffices to show that $S = 1$, where $S = \sum_{f \in \text{ColAsn}} s(f)$. For ease of notation, for each $i \in [n]$ define $X_{ii}: \text{ColAsn} \rightarrow \mathbb{F}_2$ to be the function that is identically $1$; i.e. $X_{ii}(f) = 1$ for all $i$ and for all $f \in \text{ColAsn}$. Then we may write \[S = \sum_{g \in [n]^{[n]}} \sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f)\]where $[n]^{[n]}$ denotes the set of all functions from $[n]$ to $[n]$. \[(X_{11}(f)+\boxed{X_{12}(f)}+X_{13}(f)+X_{14}(f)+\cdots)\]\[(X_{21}(f) + X_{22}(f)+X_{23}(f)+\boxed{X_{24}(f)}+\cdots)\]\[(\boxed{X_{31}(f)}+X_{32}(f)+X_{33}(f)+X_{34}(f)+\cdots)\]\[\vdots\] Figure 7. The product expression for $s(f)$ is expanded. Choose one term from each row. If $g \in [n]^{[n]}$ is not bijective, then it is not surjective (since domain and codomain have the same finite cardinality). Let $k \in [n]$ be such that $k$ is not in the image of $g$. [asy][asy] unitsize(2cm); pair a0 = (1,0)/sqrt(3); pair a1 = (-0.5,Sin(120))/sqrt(3); pair a2 = (-0.5,Sin(240))/sqrt(3); pair a10 = a1+(Cos(110),Sin(110)); label("$k$",a10+0.1*(Cos(110),Sin(110))); pair a00 = a0+(Cos(40),Sin(40)); pair a000 = a00 + (Cos(-20),Sin(-20)); pair a001 = a00 + (Cos(120),Sin(120)); draw(a0--a1, arrow = Arrow); draw(a1--a2, arrow = Arrow); draw(a2--a0, arrow = Arrow); draw(a10--a1, arrow = Arrow); draw(a00--a0, arrow = Arrow); draw(a000--a00, arrow = Arrow); draw(a001--a00, arrow = Arrow); pair rs = (4,0); pair b0 = (1,0)/sqrt(2)+rs; //dot(b0); pair b1 = (0,1)/sqrt(2)+rs; pair b2 = (-1,0)/sqrt(2)+rs; pair b3 = (0,-1)/sqrt(2)+rs; pair b10 = b1+(0,1); pair b30 = b3 + (0,-1); draw(b0--b1, arrow = Arrow); draw(b1--b2, arrow = Arrow); draw(b2--b3, arrow = Arrow); draw(b3--b0, arrow = Arrow); draw(b10--b1, arrow = Arrow); draw(b30--b3, arrow = Arrow); [/asy][/asy] Figure 8. A non-bijective function $g \in [n]^{[n]}$. Observe that the only term involving $f(k)$ in the expression for $S$ is the term corresponding to $i=k$. Since $g(k) \not= k$, we can sum over all triplets $\{f_1,f_2,f_3\}$ of color-assignments which differ only at $k$. By assumption, the vertex $(g(k),f(g(k)))$ has even degree in the induced subgraph $G_{kg(k)}$ (here we use the hypothesis that $G$ is induced-Eulerian!), so $\sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f) = 0$ if $g$ is not bijective. This implies $S$ is equal to the sum over all permutations $g$ of $\sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f)$. [asy][asy] unitsize(2cm); pair hsp = (0.3,0); pair vl = (0,-0.15); real eps = 0.1; real r = 0.05; pair a0 = (0,0); pair a1 = a0+hsp; pair a2 = a1 + hsp; pair a3 = a2 + hsp; pair a4 = a3 + hsp; pair b0 = (0,1); pair b1 = b0 + hsp; pair b2 = b1 + hsp; pair b3 = b2 + hsp; pair b4 = b3 + hsp; dot(a0); dot(a1); dot(a2); dot(a3); dot(a4); dot(b0); dot(b1); dot(b2); dot(b3); dot(b4); label("$1$",a0+vl); label("$0$",a1+vl); label("$0$",a2+vl); label("$1$",a3+vl); label("$0$",a4+vl); label("$1+0+0+1+0=0$",a2+3*vl); pair hl = (-0.8,0); label("$(g(k),-)$",(0,1)+hl); label("$(k,-)$",hl); draw(circle(b1,r)); draw(b1--a0); draw(b1--a3); draw(b0--(1-eps)*b0+eps*a1); draw(b0--(1-eps)*b0+eps*a2); draw(b0--(1-eps)*b0+eps*a3); draw(b0--(1-eps)*b0+eps*a4); draw(b2--(1-eps)*b2+eps*a0); draw(b2--(1-eps)*b2+eps*a4); draw(b3--(1-eps)*b3+eps*a0); draw(b3--(1-eps)*b3+eps*a1); draw(b3--(1-eps)*b3+eps*a2); draw(b3--(1-eps)*b3+eps*a4); draw(b4--(1-eps)*b4+eps*a0); draw(b4--(1-eps)*b4+eps*a4); draw(eps*b0+(1-eps)*a1--a1); draw(eps*b0+(1-eps)*a2--a2); draw(eps*b0+(1-eps)*a3--a3); draw(eps*b0+(1-eps)*a4--a4); draw(eps*b2+(1-eps)*a0--a0); draw(eps*b2+(1-eps)*a4--a4); draw(eps*b3+(1-eps)*a0--a0); draw(eps*b3+(1-eps)*a1--a1); draw(eps*b3+(1-eps)*a2--a2); draw(eps*b3+(1-eps)*a4--a4); draw(eps*b4+(1-eps)*a0--a0); draw(eps*b4+(1-eps)*a4--a4); [/asy][/asy] Figure 9. The induced subgraph $G_{kg(k)}$ is shown (for the sake of illustration, we have replaced the triangles with pentagons). There are $5$ ways to circle a vertex in the lower row. Summing $X_{kg(k)}$ over those five ways yields $1+0+0+1+0$, which is an even number because the circled vertex in the upper row has even degree in $G_{kg(k)}$ by assumption. If $g \in [n]^{[n]}$ is a permutation which isn't an involution, then $g^{-1}$ is a permutation which is distinct from $g$. Since $X_{ij} = X_{ji}$, we have $\sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f) = \sum_{f \in \text{ColAsn}} \prod_i X_{ig^{-1}(i)}(f)$. We can pair up the non-involution permutations into pairs, and the value of $\sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f)$ is constant within each pair. Thus the sum of $\sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f)$ over each pair is the sum of two equal numbers, which is $0$. This implies $S$ is equal to the sum over all involutions $g$ of $\sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f)$. If $g \in [n]^{[n]}$ is an involution which isn't the identity, then find $k_1, k_2 \in [n]$ such that $g(k_1) = k_2$ and $k_1 \not= k_2$. Then $\sum_{f \in \text{ColAsn}} X_{k_1k_2}(f) = 0$ since the induced subgraph $G_{k_1k_2}$ has an even number of edges (again, here we use the hypothesis that $G$ is induced-Eulerian!). This means that $\sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f) = 0$ if $g$ is an involution which isn't the identity. This implies $S$ is equal to $\sum_{f \in \text{ColAsn}} \prod_i X_{ii}(f)$. Finally, if $g$ is the identity then obviously $\sum_{f \in \text{ColAsn}} \prod_i X_{ig(i)}(f) = \sum_{f \in \text{ColAsn}} 1 = 3^n = 1$. Thus $S = 1$, as desired. $\blacksquare$