Possibilities and combinations

Question

A computer code is made up of three letters followed by four digits.

1. In how many ways is the code possible?
2. If 97 of the three-letter combinations cannot be used because they are offensive, how many codes are still possible?

For 1), I used the fundamental principle of counting:

$26\cdot26\cdot26\cdot10\cdot10\cdot10\cdot10 = 17576000$

For 2), I subtracted the total number of offensive combinations from the result I got from 1):

$17576000 - 97\cdot10\cdot10\cdot10\cdot10 = 16606000$

However, my textbook says that the answer to the second question is $174790000$.

I am wondering which answer is correct...

Answer 1

I think you did a mistake while caculating for the part 1) Answer should be 175760000 and Now if you subtract 970000 from it you will get the same answer as in your textbook.which is equal to 174790000

Answer 2

I'll assume that we are talking about 26 English-alphabet letters, and 10 decimal digits. I'm gonna call the 3-letter part the "prefix", and the 4-number part the "suffix".

Suppose we already knew that there were $m$ possibilities for the prefix and $n$ possibilities for the suffix. Then the total number of possibilities for the whole code would be $mn$.

1. If all combinations are possible, then $m=26^3$ and $n=10^4$, so $mn = 26^3\cdot 10^4 = 175,760,000$ possibilities.
2. If 97 prefixes are offensive and thus not permitted, then $m=26^3-97$ and $n=10^4$, so $mn = (26^3-97)\cdot 10^4 = 174,790,000$ possibilities.