enumeration of the number of spanning trees in some ... - Toubkal

4.3. Main Results 75Proof: It is the same proof as that of Theorem 4.3.11 with |T 1 | = τ(C.p) and |T 2 | =kτ(C − p) since there are k ways to cut the path (v 1 , v 2 , ..., v k , v k+1 ), then τ(C) =τ(C.p) + kτ(C − p).Theorem 4.3.21 Let C be a map, then|E C |1 ∑τ(C) =τ(C.e i ), where e i ∈ E C .|V C | − 1i=1Proof: Let C a map, T a spanning tree of C and e i an edge of T (there are |V C | − 1edges in T ).If in C, we look for all spanning trees that must pass through e i we find τ(C.e i ) trees witha single copy of the tree T .If we sum on the e i with i = 1, ..., |E C |, we are going to have |V C | − 1 times the tree T(because the other remaining edges |F C | − 1 do not belong to T ) then□|E C |∑τ(C.e i ) = (|V C | − 1)τ(C),i=1hence the result.□Corollary 4.3.22 Let C be a map, then|E C |1 ∑τ(C) =τ(C − e i ), such that C − e i is connected.|F C | − 1i=1Proof:From Theorem 4.3.11, we have:|E C ||E∑C ||E∑C |∑τ(C) = τ(C.e i ) + τ(C − e i )i=1i=1i=1From the previous Theorem 4.3.21, we have:|E C |∑|E C |τ(C) − (|V C | − 1)τ(C) = τ(C − e i )By Euler’s formula, we obtain the result.i=1□