i need help with permutations/combinations?

Question

I've been given the following "Digits" are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

How many 3 digit strings are there..
if any digit is allowed in any place?
if the digits are all different?
that do not start with 0 but can have any digit in the other places?
that start and end with different digits?
that have two or three digits that are the same?
the sum of whose digits is divisible by 9?
that have at least one 3?
that have at least one odd digit?

2nd part

How many strings of 9 upper-case English letters are there..
*if letters can be repeated?
if no letter can be repeated?
that start with Q, if letters can be repeated?
that start with Q, if no letter can be repeated?
that start with Q and end with X, if letters can be repeated?
that start with Q and end with X, if no letter can be repeated?
that start and end with a vowel (A, E, I, O, U), if letters can be repeated, also if they can't be repeated???

Answer 1

Mathworld has the info you need to help solve these types of problems.

http://mathworld.wolfram.com/Permutation.html

I have done part 1. Part 2 is similar but you have 26 letters instead of 10 digits.

1) 10^3 = 1000
2) 10P3 = 6x(10C3 =120) = 720
3) 9x10x10 = 900
4) 10x10x9 = 900
5) 1000-720 = 280
6) I got 57 for this. triplets such as (0,1,8) have 3!= 6 permutations, triplets such as (1,1,7) have 3!/2! = 3, and triplets such as (3,3,3) and (0,0,0) have 1. 7x6 +4x3 +2x1 = 57
7) no 3s = 9^3. Therefore at least one 3 = 10^3 -9^3 = 271
8) all even = 5x5x5=125, therefore at least one odd = 1000-125 =875

Answer 2

That's a lot of questions. I'll just give a couple of hints: In part one, you're really just dealing with the numbers from 0 to 999 (so there's a thousand of them.) In the one about divisible by 9, use the fact that a number is divisible by 9 if and only if the sum of its digits is divisible by 9, so you're just looking for all the numbers from 0 to 999 that are divisible by 9. In the one about at least one odd digit, it's easier to figure out how many numbers have no odd digits (i.e. all even digits) and then subtract that from 1000.