... ˜w4˜w5˜w6˜w7,itmustbethat m = 3 and the subword to the left of the leftmost 6 in w contains two 0’s, two 1’s, and two 2’s. However, w1must consist of 0 and 1 in some order, and w2of 1 and 2, sothe two ... 1)’s in wm, one must be on the left half of sm and hence before the m in hm+1 ,and the other must be on the right half of sk−1 and hence after the k − 2inhk−1.It now remains only to show ... first, being in w2andonthelefthalf of sk−1, would be in hk−1, and there would then be an additional pair of (k − 1)’s inwk−1, which is a contradiction.the electronic journal of combinatorics...