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 $ 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 \mathbb{N} $, then $ \alpha + \beta $ is equal to :

Hint:

Use the given recursive relation to find the probability of each word. Calculate the rank of the given word to find its serial number.

Answer 1

Correct Answer: 183

This task involves determining the lexicographical rank of a specific word within a dictionary-ordered set of permutations. It also requires analyzing a probability distribution that follows a geometric progression and subsequently calculating the sum of two parameters derived from the probability expression.

Key Concepts:

1. Word Rank: The rank of a word in a dictionary is its position when all possible permutations of a given set of letters are sorted alphabetically. This calculation involves counting preceding words by fixing letters from left to right and considering permutations of remaining letters.

2. Geometric Progression (GP): The probabilities \( P(W_n) \) adhere to a GP, where \( P(W_n) = r \cdot P(W_{n-1}) \). The sum of a finite GP is given by the formula:

\[ S_N = \frac{a(r^N - 1)}{r - 1} \]

where \( a \) is the first term, \( r \) is the common ratio, and \( N \) is the number of terms.

3. Total Probability: The sum of probabilities for all possible outcomes (in this case, all 120 unique words) must equal 1.

Solution Breakdown:

Step 1: Calculate the total number of possible words.

With five distinct letters (A, B, C, D, E), the total number of unique five-letter words is the number of permutations:

\[ N = 5! = 120 \]

Step 2: Determine the rank of the word CDBEA.

The letters in alphabetical order are A, B, C, D, E. We count words lexicographically preceding CDBEA:

1. Words beginning with A: \( 4! = 24 \) words.

2. Words beginning with B: \( 4! = 24 \) words.

3. Words beginning with C:

- Words starting with CA: \( 3! = 6 \) words.

- Words starting with CB: \( 3! = 6 \) words.

4. Words beginning with CD:

- Words starting with CDA: \( 2! = 2 \) words.

5. Words beginning with CDB:

- Words starting with CDBA: \( 1! = 1 \) word (CDBAE).

The total count of words preceding CDBEA is \( 24 + 24 + 6 + 6 + 2 + 1 = 63 \). Thus, the rank of CDBEA is \( n = 63 + 1 = 64 \).

Step 3: Establish the probability distribution.

Given \( P(W_n) = 2 P(W_{n-1}) \) for \( n>1 \), this is a GP with \( P(W_1) = p_0 \) and \( P(W_n) = p_0 \cdot 2^{n-1} \). The sum of probabilities over all 120 words is 1:

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

\[ \sum_{n=1}^{120} p_0 \cdot 2^{n-1} = 1 \]

Applying the GP sum formula with \( a = p_0 \), \( r = 2 \), and \( N = 120 \):

\[ p_0 \left( \frac{2^{120} - 1}{2 - 1} \right) = 1 \implies p_0 = \frac{1}{2^{120} - 1} \]

Step 4: Compute the probability of the word CDBEA, \( P(W_{64}) \).

\[ P(W_{64}) = p_0 \cdot 2^{64-1} = p_0 \cdot 2^{63} \]

Substituting \( p_0 \):

\[ P(CDBEA) = \frac{1}{2^{120} - 1} \cdot 2^{63} = \frac{2^{63}}{2^{120} - 1} \]

Final Calculation & Outcome:

The problem states \( P(CDBEA) = \frac{2^\alpha}{2^\beta - 1} \). By comparing this with our derived probability:

\[ \frac{2^{63}}{2^{120} - 1} = \frac{2^\alpha}{2^\beta - 1} \]

This yields:

\[ \alpha = 63 \quad \text{and} \quad \beta = 120 \]

The required sum is \( \alpha + \beta \).

\[ \alpha + \beta = 63 + 120 = 183 \]

The final value of \( \alpha + \beta \) is 183.