Probabilities: k out of n I have 10 numbers. each of them represents a coordinate. 
I think that by combining these ten numbers, 100 points can be generated:
$$10^2=100$$ 
Then by choosing k points out of those 100, there can be: $$\frac{100!}{k!(100-k)!}$$
different combinations of k points.
Is this right?
For the different values of k $$ 1<k<11$$ how many combinations can be generated?
 A: Short answer: You are correct.
Longer-ish answer (for future readers):
You have a set of ten numbers.  When you "combine" them to form coordinates, there are $10$ ways to choose the first coordinate and $10$ ways to choose the second coordinate.  Thus, there are $10\times10=100$ possible coordinates.
To select $k$ points, we use the binomial coefficient, with $n = 100$:
$$\text{Ways for k points} = \binom{n}{k} = \binom{100}{k} = \frac{100!}{k!(100-k!)}$$ 
A: Yes.
Long answer:
First, I think I'll clearly define your question. Say you mean you have a set $X_1$ with 10 distinct elements, and an equal set $X_2$ with exactly the same elements.
You can then have a third set $X_3$ which is made up of the combination of elements in the set $X_1$ and $X_2$ in that order. So if your set was $X_1$ = {a,b,c,d,e,f,g,h,i,j} which is the same as $X_2$, then $X_3$ = {aa,ab,ac,ad,ae,...jh,ji,jj}
I did that because you said "coordinates" which imply a spatial nature, which is continuous. Anyways, that doesn't really affect the answer to the question. But it does give a more defined basis.
So the size of $X_3$ is 100. If we wanted to take $k$ things from $X_3$, say $k=3$, we can take {aa,ab,ac}. Or we can take {ab,ac,ad}. Or we can take {ac,ad,ae}.
But if we take {ab,ac,aa} or {aa,ac,ab} or {ab,aa,ac} or {ac,aa,ab} or {ac,ab,aa}, that's the same as the first one we already picked, but jumbled around in a different order. The same goes for the other sets. In this case, there are 6 ways to choose the same set.
Let's try going for $k=4$. Let's pick something easy, like {aa,bb,cc,dd}.
I can pick the following sets and choose exactly the same set as above:
{aa,bb,dd,cc}
{aa,cc,bb,dd}
{aa,cc,dd,bb}
{aa,dd,bb,cc}
{aa,dd,cc,bb}
{bb,aa,cc,dd}
{bb,aa,dd,cc}
{cc,aa,bb,dd}
{cc,aa,dd,bb}
{dd,aa,bb,cc}
{dd,aa,cc,bb}
{bb,cc,aa,dd}
{bb,dd,aa,cc}
{cc,bb,aa,dd}
{cc,dd,aa,bb}
{dd,bb,aa,cc}
{dd,cc,aa,bb}
{bb,cc,dd,aa}
{bb,dd,cc,aa}
{cc,bb,dd,aa}
{cc,dd,bb,aa}
{dd,bb,cc,aa}
{dd,cc,bb,aa}
If you count them all, that's a total of 24 sets that have the same elements, but the elements have been jumbled up in a different order.
If I chose $k=5$, there would be 120 ways to pick the same set. Therefore, I won't list all of those down.
But if I chose only $k=2$, I would have only two ways to pick them out. If I chose {aa,bb}, the only other set the same as that is {bb,aa}. And if $k=1$, there's only 1 way to choose one element from that set.
So notice that when I pick $k$ things, there are a $k$! number of ways to choose the same set I chose. If I picked $k=1$, $k$!=1, and $k=2$, $k$!=2, and $k=3$, $k$!=6, and $k=4$, $k$!=24, and $k=5$, $k$!=120, and so on.
This is why your fraction is $\frac {100!}{k!(100-k)!}$. That $k$! is there in the bottom because you're removing all the other ways that you can choose essentially the same set.
This is called a combination, which you did mention. But this is the logic behind the combinations.
There is $\frac{100!}{(100-k)!}$ though, because of permutations. But as the question only wants combinations, I should probably stop here. :) It's also a good exercise for yourself to figure out why permutations work that way.
Edit:
For your follow up question, you only need to use a calculator to find out the answer yourself :)
