"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 $\mathrm{W}_{\mathrm{n}}$. Let the probability $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)$ of choosing the word $\mathrm{W}_{\mathrm{n}}$ satisfy $\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n}1$. If $\mathrm{P}(\mathrm{CDBEA})=\frac{2^\alpha}{2^\beta-1}, \alpha, \beta \in \mathbb{N}$, then $\alpha+\beta$ is equal to :____________"

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Firstly, by this rule, we note:

$ P(W_1) = p $

$ P(W_2) = 2p $

$ P(W_3) = 4p $

…

$ P(W_n) = 2^{n-1}p $

To find the initial probability $ p $, consider the total probability must sum to 1 across all 120 possible words (since $ 5! = 120 $):

$ \sum_{n=1}^{120} P(W_n) = 1 $

This is a geometric series sum where:

$ p(1 + 2 + 2^2 + \ldots + 2^{119}) = 1 $

Since the series sum $ 1 + 2 + 2^2 + \ldots + 2^{119} $ is equal to $ 2^{120} - 1 $, we have:

$ p(2^{120} - 1) = 1 \Rightarrow p = \frac{1}{2^{120} - 1} $

Thus, the probability for the $ n $-th word is:

$ P(W_n) = \frac{2^{n-1}}{2^{120} - 1} \quad \text{(i)} $

Next, determine the position of "CDBEA". Starting from the first letter:

Words beginning with 'A': $ 4! = 24 $

Words beginning with 'B': $ 4! = 24 $

Words beginning with 'C':

CA*: $ 3! = 6 $

CB*: $ 3! = 6 $

CDA$ **: 2! = 2 $

CDBA*: $ 1! = 1 $

Summing these, the position of "CDBEA" is the 64th word.

Substitute into equation (i):

$ P(\text{CDBEA}) = P(W_{64}) = \frac{2^{63}}{2^{120} - 1} $

Given $ P(\text{CDBEA}) = \frac{2^\alpha}{2^\beta - 1} $, we find:

$ \alpha = 63 $

$ \beta = 120 $

Thus, the sum $ \alpha + \beta = 63 + 120 = 183 $.

"

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