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 $$W_n$$. Let the probability $$P(W_n)$$ of choosing the word $$W_n$$ satisfy $$P(W_n) = 2P(W_{n-1})$$, $$n \gt 1$$. If $$P(CDBEA) = \frac{2^{\alpha}}{2^{\beta} - 1}$$, $$\alpha, \beta \in \mathbb{N}$$, then $$\alpha + \beta$$ is equal to __________.

There are $$5! = 120$$ different five-letter words that can be formed by using each of the letters A, B, C, D, E exactly once. These words are arranged in ordinary English (lexicographic) order and numbered $$W_1, W_2,\,\dots ,W_{120}$$.

Case 1: Determining the serial number of $$CDBEA$$

Stepwise counting in lexicographic order:

• First letter
 A-words: $$4! = 24$$ words  (serial numbers $$1\!-\!24$$)
 B-words: $$4! = 24$$ words  (serial numbers $$25\!-\!48$$)
 So words beginning with C start at serial $$49$$.

• First two letters $$C\_x$$ (remaining letters A, B, D, E):
 CA-words: $$3! = 6$$ words  (serial $$49\!-\!54$$)
 CB-words: $$3! = 6$$ words  (serial $$55\!-\!60$$)
 CD-words therefore start at serial $$61$$.

• First three letters $$CD\_x$$ (remaining letters A, B, E):
 CDA-words: $$2! = 2$$ words  (serial $$61, 62$$)
 CDB-words therefore start at serial $$63$$.

• First four letters $$CDB\_x$$ (remaining letters A, E):
 CDBA-word: $$1$$ word  (serial $$63$$)
 CDBE-words therefore start at serial $$64$$.

The only CDBE-word is $$CDBEA$$ itself, so
$$n = 64.$$

Case 2: Writing the probability model

Let $$P(W_1)=p.$$ Given $$P(W_n)=2P(W_{n-1})$$ for $$n \gt 1$$, the probabilities form a geometric progression: $$P(W_n)=2^{\,n-1}p.$$

The total probability equals $$1$$: $$\sum_{n=1}^{120}P(W_n)=p\sum_{k=0}^{119}2^k =p\,(2^{120}-1)=1.$$ Hence $$p=\frac{1}{2^{120}-1}.$$

Case 3: Probability of the word $$CDBEA$$

The required probability is $$P(CDBEA)=P(W_{64})=2^{64-1}\,p =\frac{2^{63}}{2^{120}-1}.$$

Comparing with $$P(CDBEA)=\dfrac{2^{\alpha}}{2^{\beta}-1}$$ gives $$\alpha = 63,\quad \beta = 120.$$ Therefore, $$\alpha + \beta = 63 + 120 = 183.$$

Final answer: $$183.$$