Question 636456
Your formula does not sound right. I cannot figure out how you derived it. My logic gives me different answers. (Disclaimer: I have been wrong at times)
 
I see two possible interpretations of the problem:
 
{{{highlight(IF)}}} a computer code/password is made up of two letters together, followed by 3 numbers,
like {{{highlight(AB123)}}},
you have 26 choices for the first character.
For each of those 26 choices, you have 25 choices for the second character.
That gives you 26*25 choices for the two beginning characters (letters).
For each of those 26*25 choices, you have 10 choices for the third character.
That results in 26*25*10 choices for the first three characters.
For each of those 26*25*10 choices, you have 9 choices for the fourth character.
You end up with 26*25*10*9 choices for the first four characters.
For each of those 26*25*10*9 choices, you have 8 choices for the fifth character.
That gives you {{{highlight(26*25*10*9*8)}}} choices overall.
 
However, {{{highlight(IF)}}} the letters and numbers could be scrambled into any order,
like {{{highlight(A1B23)}}}, or {{{highlight(3A2B1)}}},
there are more possibilities.
In that case, the 26*25 two letter codes, would be {{{26*25/2}}} two-letter sets,
because codes AB and BA use the same set {A, B}.
Similarly, the 10*9*8 three-digit codes represent {{{10*9*8/3/2}}} three three-digit sets.
Then, you would have {{{(26*25/2)(10*9*8/3/2)}}} different sets made of 2 letters and three digits.
Each set could be re-arranged {{{5*4*3*2}}} different ways, giving you
{{{(26*25/2)(10*9*8/3/2)(5*4*3*2)=26*25*10*9*8*5*2=highlight(26*25*10*9*8*10)}}}  different codes.