Please do not post meaningless nonsense on sciforums.
The following may be a possible path.
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
Code:
For a=1 to 26
For b=1 to 26
Print c
Next b
Next aDue 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
