An identification code is to consist of 3 letters followed by 2 digits. determine the following
a) how many different codes are possible if repetition is not permitted?
my answer is 1,263,600.
i think it is wrong. can someone me out ?

Well let's see:

The first letter can be any one of 26 .
       For each one . . .
The second letter can be any one of the remaining 25.
       For each one . . .
The third letter can be any one of the remaining 24.
       For each one . . .
The two digits can be any number from 01 to 98 ...
       except 11, 22, 33, 44, 55, 66, 77, or 88.  (No repetition.)
      There are 90 of them.

So the total number of possibilities is    (26 · 25 · 24 · 90) .

When I multiply that out, I get   1,404,000 .

I don't know how you got your number, so I can't comment on your
method, but I did find something interesting about your number:

If I assume that you did the three letters the same way I did, then
if I divide your number by  (26·25·24), the quotient will show me
how you handled the two digits.

                         1,263,600 / (26·25·24)  =  81 .

That's very intriguing, because it's so close to the 90 sets of digits
that I used.  But I don't know what it means, or if it means anything
at all.

The answer would be $68.04