I will write C(n,k) to indicate the binomial coefficient indexed by n and k (namely "n choose k"). Denote (1/n) C(n,k) C(n,k-1) by N(n,k). Prove that N(n,1)+N(n,2)+...+N(n,n) = (1/(n+1)) C(2n,n). (N(n,k) is called a Narayana number. You've just proven that the sum of the Narayana numbers is a Catalan number.)
Draw the graph whose vertices are the elements of T₃ and whose edges are the local moves that start with an edge of the form e(i,i+1). Do the same for T₄. Are the graphs T₃ and T₄ still connected if we only consider those edges?