Step-by-step explanation
Counting Possibilities with Repetition or No Repetition:
Counting the number of possibilities in a given situation can vary quite a bit depending on whether repetition is allowed or not. For example, if trying to find how many 6 letter passwords can be made with 10 letter choices, if the letters are allowed to be repeated then there are 10 choices for each letter in the password - resulting in... 10 x 10 x 10 x 10 x 10 x 10 = 10^6 =1,000,000 possible passwords.
However, if repetition of letters is not allowed, then there are 10 choices for the first letter, but only 9 for the second, 8 for the third, and so on. This results in only 10 x 9 x 8 x 7 x 6 x 5 = 151,200
Answer :
There are 26 letters to choose from and 10 digits (0 through 9).
(a) If the first three entries must all be the same letter, then there are 26 choices for the first entry, but only one choice for the second and third entries. The fourth entry can then be any letter, so there are 26 possible choices for the fourth entry. The fifth and sixth entry must be digits, but repetition of numbers are not permitted. So the fifth entry has 10 choices and the sixth entry only has 9 choices. This is a total of 26 x 1 x 1 x 26 x 10 x 9= 60,840
b) If the first letter must be one of A, B, C, or D, then there are 4 choices for the first entry. No repetition is allowed, so the second entry must be chosen out of the 25 remaining letters, and similarly the third and fourth entry will have 24 and 23 choices each. The fifth and sixth entry must be digits, but repetition of numbers are not permitted. So the fifth entry has 10 choices and the sixth entry only has 9 choices. 4 x 25 x 24 x 23 x 10 x 9=4,968,000 different codes
hope this helped alittle ..