JEE Main 2025 Apr 3 Shift 1, Mathematics Q21: Permutation

JEE Main2025Apr 3, Shift 1Mathematics

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 : ______

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Solution

There are $5! = 120$ words. With $P (W_{n}) = 2 P (W_{n - 1})$ , $P (W_{n}) = 2^{n - 1} P (W_{1}) .$

Total probability:

$\sum_{n = 1}^{120} P (W_{n}) = P (W_{1}) (1 + 2 + 2^{2} + \dots + 2^{119}) = P (W_{1}) (2^{120} - 1) = 1.$

So $P (W_{1}) = \frac{1}{2 ^{120} - 1} .$

Now find the dictionary rank of CDBEA. Count words alphabetically earlier:

A_ _ _ _: $4! = 24.$ B_ _ _ _: $24.$ CA_ _ _: $3! = 6.$ CB_ _ _: $6.$ CDA_ _: $2! = 2.$ CDBAE: $1.$ Then CDBEA itself.

Rank $= 24 + 24 + 6 + 6 + 2 + 1 + 1 = 64.$

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

Comparing with $\frac{2 ^{α}}{2 ^{β} - 1}$ : $α = 63, β = 120.$ Sum $= 183.$

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JEE Main 2025 Apr 3 Shift 1, Mathematics Q21: Permutation

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