Question

The numbers 1, 2, 3, 4,5, 6, 7, 8, 9 are arranged in a list so that each number is either greater than all the numbers that come before it or is less than all the numbers that come before it. For example, 4, 5, 6, 3, 2, 7, 1, 8, 9 is one such list: Notice that (for instance) the 6 is greater than all the numbers that come before it, and the 2 is less than all the numbers that come before it. How many such lists of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are possible?

Answer 1

The lists of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 that are possible is; 256

How to Solve Probability Combinations?

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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