Answer:
[tex]993,\!600[/tex].
Step-by-step explanation:
There is one way to choose the first character (has to be an [tex]\verb!A![/tex].)
There is one way to choose the last character (has to be a [tex]\verb!1![/tex].)
Since repetition among the letters is not allowed and the letter [tex]\verb!A![/tex] was already reserved for the first character, there would be [tex](26 - 1) = 25[/tex] English alphabet letters available for the second character.
Likewise, with the first and second characters chosen, the third character could be chosen from [tex](26 - 2) = 24[/tex] letters. There would be [tex](26 - 3) = 23[/tex] choices for the fourth letter.
Repetition of the digits is not allowed, either. With the digit [tex]\verb!1![/tex] already reserved for the last character, there would be [tex](10 - 1) = 9[/tex] ways to choose the fifth character (a digit.) There would then be [tex](10 - 2) = 8[/tex] ways to choose the sixth character (also a digit.)
Overall, the number of unique combinations would be:
[tex]\begin{aligned}& 1 \times 1 \times (25 \times 24 \times 23) \times (9 \times 8) \\ =& \; 993,\!600 \end{aligned}[/tex].
