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 by $\mathrm{W}_{\mathrm{n}}$

Accepted Answer

Firstly, by this rule, we note:

$P({W}_{1})=p$

$P({W}_{2})=2p$

$P({W}_{3})=4p$

\ldots

$P({W}_{n})={2}^{n-1}p$

To find the initial probability $p$, consider the total probability must sum to 1 across all 120 possible words (since $5!=120$):

$\sum _{120}^{n=1}P({W}_{n})=1$

This is a geometric series sum where:

$p(1+2+{2}^{2}+\ldots +{2}^{119})=1$

Since the series sum $1+2+{2}^{2}+\ldots +{2}^{119}$ is equal to ${2}^{120}-1$, we have:

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

Thus, the probability for the $n$-th word is:

$P({W}_{n})=\frac{{2}^{n-1}}{{2}^{120}-1}\,$

Next, determine the position of "CDBEA". Starting from the first letter:

Words beginning with 'A': $4!=24$

Words beginning with 'B': $4!=24$

Words beginning with 'C':

CA*: $3!=6$

CB*: $3!=6$

CDA$\ast \ast :2!=2$

CDBA*: $1!=1$

Summing these, the position of "CDBEA" is the 64th word.

Substitute into equation (i):

$P()=P({W}_{64})=\frac{{2}^{63}}{{2}^{120}-1}$

Given $P()=\frac{{2}^{\alpha }}{{2}^{\beta }-1}$, we find:

$\alpha =63$

$\beta =120$

Thus, the sum $\alpha +\beta =63+120=183$.

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