John is choosing a password for his access to the internet. He decided not to use the digit 0 or the

letter M. Each letter or number may be used more than once. How many passwords of 2 letters followed
by 4 digits are possible?
A.3,100,625
B.3,100,500
C.4,100,625
D.4,200,625

25 letters allowed for first spot

25 for second spot
so
625 letter combinations

9 digits
9*9*9*9 = 9^4
so
625*9^4 = 4,100,625

To find the number of passwords of 2 letters followed by 4 digits, we need to determine the number of possible choices for each position.

Let's break it down:

For the first letter, John can choose from any letter of the alphabet except 'M', i.e., 26 - 1 = 25 possibilities.

Similarly, for the second letter, John can choose from any letter of the alphabet except 'M', i.e., 25 possibilities.

For each of the four digits, John can choose from any of the ten numbers, i.e., 10 possibilities for each digit.

Since each position is independent, we can multiply the number of possibilities for each position to find the total number of passwords:

Total number of passwords = (Number of possibilities for first letter) * (Number of possibilities for second letter) * (Number of possibilities for the first digit) * (Number of possibilities for the second digit) * (Number of possibilities for the third digit) * (Number of possibilities for the fourth digit)

Total number of passwords = 25 * 25 * 10 * 10 * 10 * 10 = 3,100,000

Therefore, the correct answer is B.3,100,500.