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

The problem involves finding the rank of a word in a dictionary-style arrangement, analyzing a probability distribution defined by a geometric progression, and then determining the sum of two parameters from the resulting probability expression.

Concept Used:

1. Rank of a Word: The rank of a word in a dictionary is found by counting the number of words that lexicographically precede it. This is a permutation-based calculation where we fix letters in positions from left to right and count the possible arrangements of the remaining letters.

2. Geometric Progression (GP): The probabilities \( P(W_n) \) follow a GP, where \( P(W_n) = r \cdot P(W_{n-1}) \). The sum of a finite GP with first term \( a \), common ratio \( r \), and \( N \) terms is given by:

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

3. Total Probability: The sum of probabilities of all possible outcomes (all 120 words) must be equal to 1.

Step-by-Step Solution:

Step 1: Determine the total number of words possible.

The five distinct letters are A, B, C, D, E. The total number of five-letter words that can be formed by arranging these letters is:

\[ N = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

Step 2: Find the rank (serial number \(n\)) of the word CDBEA.

The letters in alphabetical order are A, B, C, D, E. We count the number of words that come before CDBEA in the dictionary.

1. Words starting with A: The remaining 4 letters (B, C, D, E) can be arranged in \( 4! = 24 \) ways.

2. Words starting with B: The remaining 4 letters (A, C, D, E) can be arranged in \( 4! = 24 \) ways.

3. Words starting with C: The first letter matches. We move to the second position. The available letters are A, B, D, E.

- Words starting with CA: The remaining 3 letters can be arranged in \( 3! = 6 \) ways.

- Words starting with CB: The remaining 3 letters can be arranged in \( 3! = 6 \) ways.

4. Words starting with CD: The second letter matches. We move to the third position. The available letters are A, B, E.

- Words starting with CDA: The remaining 2 letters can be arranged in \( 2! = 2 \) ways.

5. Words starting with CDB: The third letter matches. We move to the fourth position. The available letters are A, E.

- Words starting with CDBA: The last letter must be E. This gives the word CDBAE. There is \( 1! = 1 \) such word.

The next word in sequence is CDBEA.

The number of words before CDBEA is the sum of the counts above: \( 24 + 24 + 6 + 6 + 2 + 1 = 63 \). Therefore, the rank of the word CDBEA is \( n = 63 + 1 = 64 \).

Step 3: Determine the probability distribution.

We are given \( P(W_n) = 2 P(W_{n-1}) \) for \( n > 1 \). This defines a geometric progression. Let \( P(W_1) = p_0 \). Then \( P(W_n) = p_0 \cdot 2^{n-1} \). The sum of all probabilities must be 1:

\[ \sum_{n=1}^{120} P(W_n) = 1 \] \[ \sum_{n=1}^{120} p_0 \cdot 2^{n-1} = 1 \]

This is a GP with first term \( a = p_0 \), common ratio \( r = 2 \), and \( 120 \) terms. Using the sum formula:

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

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

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

Substituting the value of \( p_0 \):

\[ P(CDBEA) = P(W_{64}) = \frac{1}{2^{120} - 1} \cdot 2^{63} = \frac{2^{63}}{2^{120} - 1} \]

Final Computation & Result:

We are given that \( P(CDBEA) = \frac{2^\alpha}{2^\beta - 1} \). Comparing this with our calculated value:

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

By direct comparison, we find:

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

The problem asks for the value of \( \alpha + \beta \).

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

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

Answer 2

Let \( P(W_1) = x \). Given \( P(W_n) = 2P(W_{n-1}) \). Then, \( P(W_2) = 2x \), \( P(W_3) = 2^2x \), ..., \( P(W_n) = 2^{n-1}x \). Since \( \sum_{i=1}^{120} P(W_i) = 1 \), we have \[ x + 2x + 2^2x + \dots + 2^{119}x = 1 \] \[ x(1 + 2 + 2^2 + \dots + 2^{119}) = 1 \] \[ x \cdot \frac{2^{120} - 1}{2 - 1} = 1 \] \[ x(2^{120} - 1) = 1 \] \[ x = \frac{1}{2^{120} - 1} \tag{1} \] Now, let's find the rank of CDBEA. A\_ \_ \_ \_ = \( 4! = 24 \) B = \( 4! = 24 \) CA = \( 3! = 6 \) CB = \( 3! = 6 \) CDA = \( 2! = 2 \) CDBAE = 1 CDBEA = 1 So, the rank of CDBEA is \[ 24 + 24 + 6 + 6 + 2 + 1 = 63 \] 
Thus, CDBEA is \( W_{64} \). Therefore, \[ P(CDBEA) = P(W_{64}) = 2^{63} \cdot P(W_1) = 2^{63} \cdot \frac{1}{2^{120} - 1} \] \[ P(CDBEA) = \frac{2^{63}}{2^{120} - 1} \] Comparing with \( P(CDBEA) = \frac{2^\alpha}{2^\beta - 1} \), we have: \[ \alpha = 63,\quad \beta = 120 \] \[ \alpha + \beta = 63 + 120 = 183 \]