answer:Let’s say the sequence is abcdef, where each letter a-f represents one the digits 1–4. Below I use the ascii codes “element of” and “not equal to” – I hope they render right. a=1 b ∈ {2,3,4} c ∈ {1,2,3,4; c≠b} There are 3 choices for b and 3 for c. So far that makes 3*3 = 9 permutations for the combo abc d is more complicated. We see that c=1 is always one of exactly three choices for c, in which case d=1 is not allowed, whereas for c≠1 then d=1 is allowed. So of the 27 abcd combos, d=1 occurs 9 times, meaning that d≠1 occurs 27–9 = 18 times. but if d=1 then e ∈ {2,3,4} so de represents 3 combos while if d=2 then e ∈ {3,4} so de represents 2 combos if d=3 then e ∈ {2,4} so ” ” 2 ” if d=4 then e ∈ {2,3} so ” ” 2 ” In other words, if d≠1 then abcd represents 18 combos. Multiply each of these by the choices for e to enumerate all abcde, which is the same as abcdef since we know f=1. So the final answer I get is (9*3 + 18*2) = 27+36 = 61 different combos, which is notably less than 81.