A password consists of 2 letters and 2 digits (0-9). The letters and digits cannot be repeated. How many passwords are possible?

my approach:

first lets choose 2 letters and 2 numbers,
that is C(26,2)*C(10,2)
= 325*45
= 14625

let's consider one of these, say, DK45
now that could be arranged in 4! ways

so the total number of passwords is
14625*4! = 351000

bobpursley, when did they introduce that 27th letter?

C=27!/25! * 10!/8! * 4!/2!2!

The last term represents the number of permuations of two letters, two numbers can be done (six ways).

A student can create a 3 digit computer password from the numbers 0 to 9. If the digits cannot be repeated, what is the probability of someone guessing the correct password?

To determine the number of possible passwords, we need to calculate the number of choices for each position in the password and then multiply these choices together.

For the first letter, we have 26 choices since there are 26 letters in the English alphabet.
After choosing the first letter, we then have 25 choices for the second letter, as we cannot repeat the letters.
For the first digit, we have 10 choices from 0 to 9.
Finally, for the second digit, we have 9 choices because we can't repeat the digits.

To calculate the total number of passwords, we multiply the number of choices for each position together:
Number of passwords = Number of choices for first letter * Number of choices for second letter * Number of choices for first digit * Number of choices for second digit

Number of passwords = 26 * 25 * 10 * 9 = 58,500

Therefore, there are 58,500 possible passwords that can be formed using these rules.