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 $P(W_{n})=2P(W_{n-1})$, $n>1$. If $P(CDBEA)=\frac{2^{\alpha}}{2^{\beta}-1}$, $\alpha,\beta\in\mathbb{N}$, then $\alpha+\beta$ is equal to
Solution:
$$P(W_n) = 2P(W_{n-1})$$
$$P(W_2) = 2P(W_1)$$
$$P(W_3) = 2^2P(W_1)$$
$$P(W_n) = 2^{n-1}P(W_1)$$
Total number of 5-letter words formed using A, B, C, D, E is $5!= 120$.
$$\sum_{i=1}^{120} P(W_i) = 1$$
$$P(W_1) + 2P(W_1) + 2^2P(W_1) + \dots + 2^{119}P(W_1) = 1$$
This is a geometric progression with $a=P(W_1)$ and $r=2$.
$$P(W_1) \left[ \frac{1(2^{120}-1)}{2-1} \right] = 1$$
$$P(W_1) = \frac{1}{2^{120}-1} \quad \text{— (1)}$$
Rank of the word CDBEA in the dictionary:
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 CDBAE = $1$
Words starting with CDBEA = $1$
$$\text{Rank} = 24 + 24 + 6 + 6 + 2 + 1 + 1 = 64$$
So, the word CDBEA is $W_{64}$.
$$P(W_{64}) = 2^{63}P(W_1)$$
Substituting $P(W_1)$ from equation (1):
$$P(W_{64}) = \frac{2^{63}}{2^{120}-1} \quad \text{— (2)}$$
Given $P(CDBEA) = \frac{2^{\alpha}}{2^{\beta}-1}$. Comparing this with equation (2):
$$\alpha = 63, \quad \beta = 120$$
$$\alpha+\beta = 63 + 120 = 183$$
Ans. 183