Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have 10 numbers. each of them represents a coordinate.

I think that by combining these ten numbers, 100 points can be generated:


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?

share|cite|improve this question
Yes, you're right. – markovchain Jun 25 '13 at 12:25

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 $n = 100$: $$\text{Ways for k points} = \binom{n}{k} = \binom{100}{k} = \frac{100!}{k!(100-k!)}$$

share|cite|improve this answer
I think you have to go even further, it works for coordinates, but in a set of 10 numbers there are actually $2^{10}=1024$ different subsets of 10 numbers. – Salieri Jun 25 '13 at 12:37
Why do you care about subsets, @Cardonai ? – Thomas Andrews Jun 25 '13 at 12:50
@Cardonai You're right: a set of $10$ numbers has a power set of size $2^{10}$. But, I don't see the bearing this has on my answer... – apnorton Jun 25 '13 at 12:51


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.


For your follow up question, you only need to use a calculator to find out the answer yourself :)

share|cite|improve this answer
OK! Many thanks!!! I am not sure that I can use a calculator because I want to use different values than 100. I mean that I want to study this function $$\frac{n!}{k!(n-k)!}$$ – Herc11 Jun 25 '13 at 12:59
As far as that goes, you can study it by observing a graph. But I recommend you study it by learning the logic of how that formula came to be. On a calculator, you can just use ${_n}C_r$. You can even do this on the internet. Go to and play with your numbers. – markovchain Jun 25 '13 at 13:28
Yes. I used Matlab. I saw that when k>50 the number of combinations is smaller than k=50. Is the same graph as binomial's distribution? – Herc11 Jun 25 '13 at 13:32
Almost the same. In fact, the binomial distribution is $$\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$$ – markovchain Jun 25 '13 at 13:36
The reason it peaks at k=50 is because $\frac{100!}{k!(100-k)!}$ when k=1 is the same as when k=99. And it is the same when k=2 and k=98. And so on, up to k=49 and k=51. Generally, ${_n}C_r$ (the combination) is equal when r=k and when r=n-k. Can you see why? – markovchain Jun 25 '13 at 13:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.