Distance coefficients are usually computed by summation over differences with respect to individual, suitably scaled characters. A problem arises in the case of a nominal (non-ordered) character of n states (n > 2). Any mismatch in character states might be regarded as just that-a mismatch-and be scored as unit difference for the character; and I suspect that this is the conventional treatment (the point is rarely mentioned explicitly). On the other hand, the attitude might be taken that unit difference is appropriate for a two-state character; but for more than two states there are correspondingly greater chances of a mismatch. Therefore, mismatches should be downweighted in some appropriate fashion. An alternative approach, which also provides a solution, is the following. A non-ordered character can be regarded as in principle ordered: simply, the information required to order its states is lacking. For instance, we may take it as given that the states did arise in some order (not necessarily linear) during the evolution of the species whose members characteristically exhibit them. The character state trees for those characters accepted as ordered are not actually known. Simply, they show a logico-mathematical structure that allows a reasonable guess at an undirected ordering (and in phenetics directed ordering is not required). For others (e.g., eye colors, irregular shapes, amino acids in protein sequences), there may be no basis for guessing-unless, of course, one is prepared to assume parsimonious evolution on a tree of given shape (Fitch, 1971; Mickevich, 1982). If, then, such a nominal character had, say, three states, its ordering could take any one of three forms (all linear). Thus, for such a character (normalized on, say, range) and any given mismatch between two OTUs, the true distance between the OTUs would be 0.5 for two of the potential orderings and 1.0 for the remaining one. Therefore, to score unit difference for every such mismatch would on average, give excessive weight to the character in two cases out of three. An obvious solution is to score the average of the scores for a given mismatch on all possible orderings (character state trees). For instance, in the foregoing example the value would be (1.0 + 0.5 + 0.5)/3 = 0.67 for all mismatches in the character concerned. In such a fashion, the score would sometimes be a little under the score and sometimes a little over; but, lacking information, we could do no better. For a character with n states there are nn-2 possible trees (linear and branched) and, with increasing n, it becomes increasingly difficult to calculate the average score. Fortunately, characters with more than five or six states are very rare. I have made the calculations empirically, by simple consideration of tree shapes and numbers (Harary, 1969), for n of 3, 4, and 5, and W. H. E. Day (pers. comm.) has done the same for an n of 6. The corresponding scores are: 0.67, 0.60, 0.56, and 0.53; and by graphical extrapolation, a score of 0.52 seems a reasonable approximation for 6 < n < 10. I am incorporating these weights into my own programs for computing distance coefficients, and would recommend them to others for trial at least. However, it remains possible that, for certain characters, it may be thought that the state tree would always be linear and branching most unlikely. For such a case, and n states, the weights can be computed directly, as (n + 1)/3(n 1).