Question

An identification code is to consist of 3 letters followed by 2 digits. Determine the following

Answer 1

Answer:

Number of codes possible = 1404000

Step-by-step explanation:

We need to find how many different codes are possible if repetition is not permitted.

Here the code consist of 3 letters and 2 numbers in order.

Consider first place,

          Here we need to choose 1 letter from 26 letters

          Number of ways = 26

Consider second place,

          Here we need to choose 1 letter from 25 letters

          Number of ways = 25

Consider third place,

          Here we need to choose 1 letter from 24 letters

          Number of ways = 24

Consider fourth place,

          Here we need to choose 1 number from 10 numbers

          Number of ways = 10

Consider fifth place,

          Here we need to choose 1 number from 9 numbers

          Number of ways = 9

So total number of ways = 26 x 25 x 24 x 10 x 9 = 1404000

Number of codes possible = 1404000

Answer 2

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.