All five letter words are ad 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 . Let the probability of choosing the word satisfy . If , then is equal to ____

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Solution

Q.
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 $P (C D B E A) = \frac{2^{α}}{2^{β} - 1}, α, β \in ℕ$ , then $α + β$ is equal to __________. [2025]

Ans.
(183)

Let $P (W_{1}) = x$

Possible arrangement of 5 letters = 5! = 120

Now, $\sum_{i = 1}^{120} P (W_{i}) = 1$

$\Rightarrow x + 2 x + 2^{2} x + 2^{3} x + . . . . . + 2^{119} x = 1$

$\Rightarrow \frac{x (2^{120} - 1)}{(2 - 1)} = 1 \Rightarrow x = \frac{1}{2^{120} - 1}$ ... (1)

Possible arrangements after fixing letters are given by

A _ _ _ = 4! = 24

B _ _ _ _ = 4! = 24

C A _ _ _ = 3! = 6

C B _ _ _ = 3! = 6

C D A _ _ = 2! = 2

C D B A E = 1

C D B E A = 1

${Rank of C D B E A = 64}_{_}^{_}$

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

On Comparing with $\frac{2^{α}}{2^{β} - 1}$ , we get $α$ = 63 and $β$ = 120.

$\Rightarrow α + β = 63 + 120 = 183$ .

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