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 Wₙ. Let the probability P(Wₙ) of choosing the word Wₙ satisfy P(Wₙ)=2 P(Wₙ₋₁), n>1.
If P(CDBEA)=2^α/(2^β-1), α, β ∈ ℕ, then α+β is equal to :____________

Solution

Maths sir · 4 steps

Find the rank of CDBEA first; the doubling rule then makes every probability a power of 2.

1. Rank of CDBEA. Words starting with A or B: 2 × 4! = 48. Starting CA or CB: 2 × 3! = 12, total 60. Starting CDA: 2! = 2, total 62. Then CDBAE is 63 and CDBEA is 64.
2. Doubling rule. Each probability is twice the one before it, so P(Wₙ) = 2ⁿ⁻¹ p, where p = P(W₁). There are 5! = 120 words, and their probabilities must add up to 1.

   p(1 + 2 + 2² + … + 2¹¹⁹) = p(2¹²⁰ - 1) = 1

3. Put in rank. So p = 1/(2¹²⁰-1), and P(CDBEA) = P(W₆₄) = 2⁶³/(2¹²⁰-1). This gives α = 63 and β = 120.
4. So. α + β = 63 + 120 = 183.

The idea to keep. Fix letters from the left one at a time, and count the words that start with a smaller letter in that place. A probability that doubles each time is a GP, so the total of 1 fixes the first term.

Answer from the NTA final answer key. Solution by PrepWiser.