How to find the pattern? I'm challenging myself to figure out the mathematical expression of the number of possible combinations for certain parameters, and frankly I have no idea how.
The rules are these:
Take numbers 1...n. Given m places, and with no repeated digits, how many combinations of those numbers can be made?
AKA


*

*1 for n=1, m=1 --> 1

*2 for n=2, m=1 --> 1, 2

*2 for n=2, m=2 --> 12, 21

*3 for n=3, m=1 --> 1,2,3

*6 for n=3, m=2 --> 12,13,21,23,31,32

*6 for n=3, m=3 --> 123,132,213,231,312,321


I cannot find a way to express the left hand value. Can you guide me in the steps to figuring this out?
 A: I believe what you are looking for is 
$P_m^n$   
Permutations so far has hold. 
The equations is $\frac{n!}{(n-k)!}$ 
where ! means factorial
 N is the set and K would be the size of the set that you select from the larger set N. 
Or in your case $\frac{n!}{(n-m)!}$ 
A: Suppose you have to construct such a sequence of length $m$ using $n$ different objects (in your case, digits).
To create such a sequence, you can first pick any of the $n$ objects as the first entry of your sequence. Then you pick the second entry, but this time you only have $n-1$ choices remaining (since you're not allowed to use the same object twice). Next, you pick the third entry, and for this you have $n-2$ choices.
Continuing in this way, you would have a total of
$$ n(n-1)(n-2)\cdots (n-m+1) $$
choices (to see that you stop at the factor $n-m+1$, realize that you need a total of $m$ factors, one for each entry you have to pick).
So the answer is
$$ n(n-1)(n-2)\cdots (n-m+1)$$
which is sometimes written $P_{n,m}$ (or $P^n_m$ or ...). For those who prefer using the notation with factorials (personally, I do, but it's easier to see where the formula comes from without factorials), this is just
$$ \frac{n!}{(n-m)!}.$$
Wikipedia has an article on permutations, including the solution to your problem (at "There is also a weaker meaning").
