All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number n be denoted by Wn. Let the probability P(Wn) of choosing the word Wn satisfy P(Wn) = 2P(Wn−1), n 1. If P(CDBEA) = P(W64) = 2^α/(

Question

All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number n be denoted by Wn. Let the probability P(Wn) of choosing the word Wn satisfy P(Wn) = 2P(Wn−1), n > 1. If P(CDBEA) = P(W64) = 2^α/(2^β − 1), α, β ∈ N, then α + β is equal to:

Accepted Answer

183. The probability of choosing the word Wn is defined such that each subsequent word's probability is double that of the previous word, indicating an exponential growth pattern. Given that P(W64) can be expressed in the form 2^α/(2^β - 1), solving for α and β based on the established pattern leads to the conclusion that their sum equals 183.

Solution

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