The order of the last three characters is not important in either exercise. The importance is knowing how many letters of the alphabet are left from which to choose for each of these three slots.
Here's a similar example.
A PIN is to be four characters, all digits. The first character must be either 4 or 5. How many PINs are possible with repitition and without repitition.
There are only ten digits (0 through 9).
WITHOUT REPITITION:
There are two choices for the first character (4 or 5).
Once that choice is made, there are nine choices available for the second character.
After that, there are eight choices remaining for the third character.
There are seven choices from which to pick the fourth character.
2 * 9 * 8 * 7 = 1008 possible PINs
WITH REPITITION:
There are two choices for the first character.
There are ten choices for the second, third, and fourth characters.
2 * 10 * 10 * 10 = 2000 possible PINs
If you google "counting problems" (with the quotes), then you will find more lessons than you can count.
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