All five-letter words are made using all the letters A, B, C, D, E and arranged as in an E…

Question

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 $W_n$. Let the probability of choosing the word $W_n$ satisfy $P(W_n) = 2P(W_{n-1})$, $n>1$. If $P(	ext{CDBEA}) = \dfrac{2^\alpha}{2^\beta-1}$, $\alpha,\beta \in \mathbb{N}$, then $\alpha+\beta$ is equal to: \_\_\_\_\_\_\_\_\_\_\_\_

Answer: 183

Accepted Answer

Since $P(W_n) = 2^{n-1}p$ where $p=P(W_1)$, and $\sum_{n=1}^{120}P(W_n)=1$:
$$p(2^{120}-1) = 1 \implies p = \frac{1}{2^{120}-1}$$

So $P(W_n) = \dfrac{2^{n-1}}{2^{120}-1}$.

Finding the position of "CDBEA": words starting with A ($4!=24$), B ($24$), then within C:
- CA: $3!=6$
- CB: $3!=6$
- CDA*: $2!=2$
- CDBA*: $1!=1$

Position $= 24+24+6+6+2+1 = 63$... rechecking against source total of $64$ (position numbering): the position works out to $64$ (i.e., CDBEA is the $64^{\text{th}}$ word).

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

So $\alpha=63,\ \beta=120$, and $\alpha+\beta = 183$.

Solutions · 1