The number of permutations, of the digits 1, 2, 3, ..., 7 without repetition, which neither contain the string 153 nor the string 2467, is _______.

We need the number of permutations of 1-7 that contain neither "153" nor "2467" as substrings. The total number of permutations is $$7! = 5040$$.

Counting those containing "153", we treat "153" as a single block, giving 5 objects to arrange: [153], 2, 4, 6, 7, so $$|A| = 5! = 120$$. Counting those containing "2467", we treat "2467" as a single block, giving 4 objects: 1, [2467], 3, 5, so $$|B| = 4! = 24$$.

When both "153" and "2467" are present, the two blocks use all 7 digits, and arranging the blocks [153] and [2467] gives $$|A \cap B| = 2! = 2$$. By inclusion-exclusion, $$|A \cup B| = 120 + 24 - 2 = 142$$, so the number of permutations without these substrings is $$5040 - 142 = 4898$$.

The answer is 4898.