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 $P(W_n) = 2P(W_{n-1})$, $n > 1$. If $P(CDBEA) = \frac{2^\alpha}{2^\beta-1}$, $\alpha, \beta\in N$, then $\alpha + \beta$ is equal to :
The problem requires finding the sum $α + β$ based on the probability distribution of five-letter words formed using {A, B, C, D, E} and arranged in dictionary order.
Step 1: Determine Total Number of Words (N)
The words are five letters long using 5 distinct letters (A, B, C, D, E). The total number of possible permutations (words) is:
N = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
Step 2: Find the Serial Number (n) of Word CDBEA
We determine the rank of the word CDBEA by counting words that precede it in dictionary order.
Letters available: {A, B, C, D, E}. Total words N = 120.
1. Contribution from first letter: Words starting with A (4! = 24 permutations) and B (4! = 24 permutations) come before any word starting with C. Total contribution = 24 + 24 = 48.
2. Contribution from second letter (given C): Remaining letters are {A, B, D, E}. Words starting with CA (3! = 6 permutations) and CB (3! = 6 permutations) come before words starting with CD. Total contribution = 6 + 6 = 12.
3. Contribution from third letter (given CD): Remaining letters are {A, B, E}. Words starting with CDA (2! = 2 permutations) come before words starting with CDB. Total contribution = 2.
4. Contribution from fourth letter (given CDB): Remaining letters are {A, E}. Words starting with CDBA (1! = 1 permutation, specifically CDBAE) come before words starting with CDBE. The target word CDBEA follows CDBAE. Total contribution = 1.
Total words before CDBEA: Summing the counts = 48 + 12 + 2 + 1 = 63.
The serial number, n, for the word CDBEA is therefore 63 + 1 = 64.
Step 3: Determine the Probability Formula
Let $Wk$ denote the k-th word in the dictionary order and $P(Wk)$ be its probability.
The given probability relation is: $P(Wn) = 2P(Wn-1)$ for $n > 1$.
This indicates a geometric progression: $P(Wk) = P(W1) \cdot 2k-1$.
The sum of probabilities for all words must be 1:
\sum_{k=1}^{N} P(W_k) = 1
Substituting the geometric progression formula:
P(W_1) \sum_{k=1}^{N} 2^{k-1} = 1
Using the formula for the sum of a geometric series (\(\sum_{i=0}^{m-1} r^i = \frac{r^m - 1}{r-1}\)), we get:
P(W_1) \left( \frac{2^N - 1}{2-1} \right) = 1 \implies P(W_1) (2^N - 1) = 1
Thus, the probability of the first word is \(P(W_1) = \frac{1}{2^N - 1}\).
The general formula for the probability of the k-th word becomes:
P(W_k) = \frac{2^{k-1}}{2^N - 1}
Step 4: Calculate P(CDBEA)
Using the derived formula with $N = 120$ and the serial number $n = 64$ for CDBEA:
P(CDBEA) = P(W_{64}) = \frac{2^{64-1}}{2^{120} - 1} = \frac{2^{63}}{2^{120} - 1}
Step 5: Find Alpha (α) and Beta (β)
The problem states the probability in the form:
P(CDBEA) = \frac{2^\alpha}{2^\beta - 1}
By comparing this with the calculated probability \(\frac{2^{63}}{2^{120} - 1}\), we identify:
- \(\alpha = 63\)
- \(\beta = 120\)
Both $α$ and $β$ are natural numbers ($α, β \in \mathbb{N}$).
Step 6: Calculate α + β
Finally, compute the sum $α + β$:
\alpha + \beta = 63 + 120 = 183
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