Maths Q22 – JEE-MAIN 2025

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}) = 2 P (W_{n - 1})

n > 1

. If

P (CDBEA) = \frac{2^{α}}{2^{β} - 1}

α, β \in N

, then

α + β

is equal to : _______

Enter Calculated Numerical Value:

Clue 1 of 3: Permutations and Lexicographical Order

First, determine the total number of possible words and the serial number of the specific word "CDBEA" when arranged lexicographically.

🧠 Full Solution Path

Step 1: Determine the serial number of CDBEA✦ Active

The letters are A, B, C, D, E. The total number of 5-letter words (permutations) is $5! = 120$ . We need to find the serial number of the word CDBEA in lexicographical order:

\begin{array}{ll} Words starting with A: & 4! = 24 \\ Words starting with B: & 4! = 24 \\ Words starting with CA: & 3! = 6 \\ Words starting with CB: & 3! = 6 \\ Words starting with CDA: & 2! = 2 \\ Words starting with CDBA: & 1! = 1 (CDBAE) \\ The next word is CDBEA. \end{array}

The serial number $n = 24 + 24 + 6 + 6 + 2 + 1 + 1 = 64$ . So, $W_{64} = CDBEA$ .

Step 2: Find the general probability

P (W_{n})

○ Expand

Given $P (W_{n}) = 2 P (W_{n - 1})$ for $n > 1$ . This implies that the probabilities form a geometric progression with a common ratio of 2. Let $P (W_{1}) = P_{1}$ . Then $P (W_{n}) = P_{1} \cdot 2^{n - 1}$ .

The sum of probabilities for all $N = 120$ words must be 1:

\sum_{n = 1}^{120} P (W_{n}) = 1 ⟹ \sum_{n = 1}^{120} P_{1} \cdot 2^{n - 1} = 1

P_{1} (1 + 2 + 2^{2} + \dots + 2^{119}) = 1

P_{1} \frac{2^{120} - 1}{2 - 1} = 1 ⟹ P_{1} = \frac{1}{2^{120} - 1}

Therefore, the general probability is $P (W_{n}) = \frac{2^{n - 1}}{2^{120} - 1}$ .

💡 Teacher's Secret Hint

Remember that the sum of all probabilities in a sample space must equal 1.

Step 3: Calculate

P (CDBEA)

and find

α + β

○ Expand

For the word CDBEA, we found its serial number to be $n = 64$ . Substitute this into the probability formula:

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

We are given that $P (CDBEA) = \frac{2^{α}}{2^{β} - 1}$ . Comparing the two expressions, we get:

α = 63 and β = 120

Both $α = 63$ and $β = 120$ are natural numbers, as required. Finally, calculate $α + β$ :

α + β = 63 + 120 = 183

💡 Teacher's Secret Hint

Ensure to correctly identify the exponent for the numerator and the base for the denominator from the derived probability expression.

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Maths Question 22 – JEE-MAIN 2025

Clue 1 of 3: Permutations and Lexicographical Order

🧠 Full Solution Path