Number of $4 $ digit numbers with no repeated digit. Number of $4$ digit numbers with no repeated digit is 


*

*$4536$

*$3024$

*$5040$

*$4823$


Well, I am very much weak in combinatorics. Please help.
 A: Ok so lets write down any old $4$ digit number
$abcd$
How many choices do we have for the digit $a$? We have $9$ choices (since the first digit cannot be $0$). Now for each possible choice of $a$ we have $9$ choices for $b$ (since we want $b$ to be a different digit to $a$ and we now allow $0$).
So for choice of the $ab$ part we have $9*9 = 81$ possibilities.
Now for each of these we have $8$ choices for $c$ (to avoid $c$ being the same as either $a$ or $b$). And for each of these we have $7$ choices for $d$ (to avoid $d$ being the same as either $a,b$ or $c$).
So in total there are $9*9*8*7 = 4536$ possible numbers.
A: A bit more hint:
Ask yourself how many ways can you choose the digit in the first position, then how many ways can you choose the second-place digit ...
A: Hint: this is like drawing four digits out of a bag of ten digits 0-9. How many different ways can this happen?
Edit: To correct the model I had in mind, the bag would first have to contain only 1-9. After you've picked the leftmost digit, you would throw in a 0 ball and continue to pick the last three digits.
A: When you say "combinations" I also get the sense you mean that order doesn't matter. Sometimes people say "combinations" when they mean "permutations" where the order matters.
So for example, if there are three people April, Bruce and Chandra and I want to know how many different pairs of people there could be there are three: April and Bruce, Bruce and Chandra, or Chandra and April. The orders don't matter because April and Bruce is the same as Bruce and April. This is a "combination" of people. Thus, the number of ways of choosing 2 people from 3 is three.
On the other hand, if those three people competed in a math club, then how many different ways could you choose a winner and a runner-up? It could be April first and Chandra second, or Chandra first and April second, or four other possibilities involving Bruce. Thus, the number of ways of choosing 2 people from 3 where order matters is six. This is working with "permutations".
The two ideas are related. If you take the number of permutations and divide it by the number of ways of arranging two people (2) then you get the number of combinations.
Now to your question:
Since you asked about combinations, I will say that order doesn't matter.
Start by finding the permutations: For the first choice, you have 10 possible digits to choose from. For the second choice, you have 9 digits because you used one for the first choice. The third choice comes from 8 possibilities and the fourth from 7 possibilities. Now we multiply these together: 10 x 9 x 8 x 7 = 90 x 56 = 5040. That's the number of permutations. No digits repeat, but 0123 is different from 0321.
Now to find the number of combinations, I have to know how many different ways there are of arranging four digits. That's the same kind of problem: the first position could be from 4 possibilities, the second from 3 possiblities, the third from 2 choices and the last has to be the 1 left. So there are 4 x 3 x 2 x 1 = 24 possible ways of arranging 4 items.
Therefore I divide 5040 / 24 = 210. So there are 210 different combinations of four digits chosen from 0-9 where the digits don't repeat.
