The lists of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 that are possible is; 256
<h3>How to Solve Probability Combinations?</h3>
We are told that the last digit has to be either a one (1) or nine (9), then are two (2) options (>,<) for the remaining eight digits, so there are 2⁸ = 256 arraignments where each digit is either greater than or less than the preceding digits. This can be confirmed as below;
Numbers beginning with 1 = 8C0 = 1
Numbers beginning with 2 = 8C1 = 8
Numbers beginning with 3 = 8C2 = 28
Numbers beginning with 4 = 8C3 = 56
Numbers beginning with 5 = 8C4 = 70
Numbers beginning with 6 = 8C5 = 56
Numbers beginning with 7 = 8C6 = 28
Numbers beginning with 8 = 8C7 = 8
Numbers beginning with 9 = 8C8 = 1
Total lists of numbers possible = 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = 256
Read more about Probability Combinations at; brainly.com/question/25688842
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