math help !!!!!!

Question

Marcus is creating a password with 2 digits followed by 4 letters. If digits and letters cannot be repeated, determine the number of different possible passwords.

Answer 1

For a password, note that the order does matter: if your password is "cat", then "tac" will not be accepted. So we're doing permutations. And in every password system that I've ever seen, passwords are case-dependent: "Cat" and "caT" are different. So we have 52 characters in the case-dependent English alphabet.

Putting that all together, we should have 10*9 permutations for the first 2 digits, followed by 52*51*50*49 permutations of 4 non-repeating characters from a 52-character alphabet, multiplied together:

10*9*52*51*50*49. Or, if you like factorials, (10! / 8!) * (52! / 48!), using the formula for the number of permutations of N things taken R at a time, where order matters and you are not allowed to use the same thing twice.

The answer: 584,766,000.

Answer 2

10 possibilities for 1st digit times 9 possibilities for 2nd digit times 26 possibilities for 1st letter times 25 possibilities for 2nd letter times 24 possibilities for 3rd letter times 23 possibilities for 4th letter gives:

10 × 9 × 26 × 25 × 24 × 23 or 32292000 possible outcomes.

Dividing 32292000 by 2! and then by 4! will give 672750 possible outcomes. This assumes that 2 digits chosen cannot be chosen again in any order and that 4 letters chosen cannot be chosen again in any order.

The statement of the problem does not clarify whether digits (and/or letters) can be chosen again in different order.

Answer 3

If the digits are 0-9 and 26 letters in the alphabet, the number of combinations without repeats should be

₁₀C₂*₂₆C₄ = 672,750