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Let S(n) be a collection of subsets of f1;:::;ng .I n this paper we study numerical obstructions to the existence of orderings of S(n )f or which the cardinalities of successive subsets satisfy congruence conditions. Gray code orders provide an example of such orderings. We say that an ordering of S(n) is a Gray code order if successive subsets dier by the adjunction or deletion of a single element of f1;::: ;ng. The cardinalities of successive subsets in a Gray code order must alternate in parity. It follows that if d(S(n)) is the dierence between the number of elements of S(n) having even (resp. odd) cardinality, then jd(S(n)) j 1 is a lower bound for the cardinality of the complement of any subset of S(n) which can be listed in Gray code order. For g 2, the collection B(n;g )o f g-blockfree subsets of f1;::: ;ngis dened to be the set of all subsets S of f1;::: ;ngsuch that ja b j gif a; b 2 S and a 6= b. We will construct a Gray code order for B(n; 2). In contrast, for g> 2 we nd the precise (positive) exponential growth rate of d(B(n; g)) with n as n !1 . This implies B(n;g) is far from being listable in Gray code order if n is large. Analogous results for other kinds of orderings of subsets of B(n; g) are proved using generalizations of d(B(n; g)) . However, we will show that for all g, one can order B(n;g) so that successive elements dier by the adjunction and/or deletion of an integer from f1;::: ;ng . We show that, over an A-letter alphabet, the words of length n which contain no block of k consecutive letters cannot, in general, be listed so that successive words dier by a single letter. However, if k> 2a nd A> 2o r if k =2 and A> 3, such a listing is always possible.

Combinatorial families that are exponentially far from being listable in Gray code sequence