Better than a computer.

The following may be a possible path.

Code:

For a=1 to 26

For b=1 to 26

Print c

Next b

Next a

The former is a two-letter word, while the, "for" loops count through the letters of the alphabet.

Due to the nesting of the loops, the first loops count through the twenty-six letters of the alphabet (26^letterposition) times. So then the words' position may be calculated as follows:

(26^letterposition)+(26^letterposition)+letterposition.

This, however, proves to be an inefficient code because the numbers involved are far too large. It is how the decimal system works though:

(1×10^3)+(2×10^2)+(3×10)+4

 

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