Permutation Problem with Given Numbers Given the digits 0,2,5,6,9.
A. How many 3-digit numbers can be formed if no two digits are to be the same?
B. Of the numbers formed, how many are even? How many are odd? How many are greater than 600?
C. How many 3-digit numbers if the digit may be repeated?
 A: When tackling these kinds of problems you need to classify the problem.  There at two types: permutations and combinations.  Permutations are when you are looking for the number of distinct elements; Combinations are when you are not looking for distinction, e.g, {1,2,3},{2,1,3},{3,1,2},{3,2,1},{1,3,2} can all be classified as {1,2,3}, that is, order does not matter. For permutations, order does matter.  Now, there are two scenarios you have to consider for permutations and combinations:  
1) repetition is allowed
2) repetition is not allowed.
In your example, for part A), you are looking for the number of distinct 3 digit numbers that can be formed without repetition.  So, since there are $5$ total elements, and you want no repetition, then the total number of permutations will be $5*4*3*2*1.$ But since you only want the first three, then you have to stop counting when you get to $2,$ that is $5*4*3.$  Then the answer will be $\dfrac{5!}{(5-3)!}=\dfrac{5!}{2!}=5*4*3=60$ distinct numbers without repetition.  This may help you to solve the other problems.  Let me know if you still need help.
A: A. How many 3-digit numbers can be formed if no two digits are to be the same?
Imagine our number has been formed as A B C.
For A we have 4 choices: 2, 5, 6, 9;
For B we have 4 choices: 0 and the 3 digits that not have been used in A;
For C we have 3 choices: digits that have not been used in A and B.
So, using the fundamental counting principle we have: 4 * 4 * 3 = 48 numbers.
