How many numbers between 4,000 and 7,000 can be chosen using the digits [0, 8]? I have a homework problem in combinatorics, and I am struggling to solve it because I didn't understand our lesson well.
Can you please help me to solve this problem?

How many numbers between 4,000 and 7,000 can be chosen using the digits 0, 1, 2, 3, 4, 5, 6, 7, and 8 if each digit must not be repeated in any numbers?

PS: I don't have the resources to solve this. I just don't understand what our professor taught us, but we have already made a make-up class to solve this conflict.
 A: There are some correct answers but they don't explain well enough for my taste. 
First, to think about these types of problems, it helps to visualize what is going on. In this case, I would use four spaces to visualize a four digit number.
$$_ _ _ _$$
Now we want to figure out how many ways we can fill in each blank, being very careful to count each possibility once and only once. First(and I am not going into full detail here) we usually want to start with the MOST restricted choice. In this case, that is the first digit. Since 7000 isn't allowed due to repetition, we can have either 4,5 or 6 as the first digit. Thus there are 3 possibilities. Now if we had a number in the first place(for the sake of visualization say 4)
$$4 _ _ _$$
Now how many choices do we have for the next digit? There are 8 allowable digits we haven't used yet, so eight ways we can fill in the second digit(did this depend on our choice of using 4 as the first digit?)
What about after that? The third digit now has seven possibilities and the fourth is left with six. Thus we can count this as 
$$3*8*7*6=1008$$
It should also be noted that this is counted by $$3*P(8,3)$$ since the last three digits can be thought of as lining up three objects from 8 since order matters, repetition is not allowed and there are no other restraints on the choices(the three things to check for before using a permutation formula)
A: HINT...you have a choice of 4, 5 or 6 for the first place (from left to right).
You then have a choice of 8 digits for the second place (including 0 but excluding your first choice).
Then you have a choice of 7 digits for the third place.....
can you finish this?
A: You first have to choose the first digit ($\neq 0,9$), then choose any $3$ distinct digits among the remaining ones, taking into account their order. This is called the number of $3$*-arrangements* on the set of remaining digits.
Let's compute all this: 
General case: the first digit can be any digit, but $0$ and $9$. Hence there are $8$ possible  choices. For the other digits you need to choose a list of $3$ distinct digits among the $8$ remaining digits. There are
$$A_8^3=\frac{8!}{(8-3)!}=8\cdot7\cdot6,\enspace\text{whence}\quad 8^2\cdot7\cdot 6\enspace\text{possibilities}.$$
Favourable case: the first digit must be one of $4,5,6$, i.e. there are $3$ possible choices. The other digits are as above, whence 
$$3\cdot8\cdot7\cdot6=1008 \enspace\text{favourable cases.}$$ 
If we choose as sample space the set of quadruples of digits between $0$  and $8$, the first digit being different from $0$,  the probability of having a number with distinct digits in $\{\,0,\dots,8\,\}$, between $4000$ and $7000$, is equal to $\dfrac38$.
Choosing as a sample space the set of all $4$-digit numbers, the answer is different for the general case: we have $9000$ possibiliites in all, so the probability is:
$$\frac{3\cdot8\cdot7\cdot6}{3^2\cdot 2^3 \cdot5^3}=\frac{14}{125}.$$
A: You got $3 (4,5,6)$ choices for the first place.
8 choices for the second place, 0-8 and excluding the one number you already chose for the first place.
7 for third one and 6 for the last.
Overall, using this you can make $3*8*7*6=1008$ such numbers.
A: The first digit can be 4,5,6 footie that we exclude 7 because 7000 repeats 0, so there are 3 possible choices for the first digit. Then you have 8 choices (0,1,2,3,4,5,6,7,8 excluding the first digit) for the next digit, and each following digit has one less. So you get 3*8*7*6=1008. So there are 1008 possible numbers.
