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 \(\mathrm{W}_{\mathrm{n}}\). Let the probability \(\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)\) of choosing the word \(\mathrm{W}_{\mathrm{n}}\) satisfy \(\mathrm{P}\left(\mathrm{W}_{\mathrm{n}}\right)=2 \mathrm{P}\left(\mathrm{W}_{\mathrm{n}-1}\right), \mathrm{n}>1\). If \(\mathrm{P}(\mathrm{CDBEA})=\frac{2^{\alpha}}{2^{\beta}-1}, \alpha, \beta \in \mathbb{N}\), then \(\alpha+\beta\) is equal to :

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Permutations and Combinations

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 :

Answer: 183

Solution

There are

5! = 120

words, and with

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

the probabilities sum to

P (W_{1}) (2^{120} - 1) = 1

, so

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

. Counting words before CDBEA in dictionary order:

24

begin with A,

24

with B,

6

with CA,

6

with CB,

2

with CDA, and

1

with CDBA, totalling

63

, so CDBEA is

W_{64}

. Therefore

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

, so

α = 63, β = 120

and

α + β = 183

JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Integer answer

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JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Permutations and Combinations

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