Number of different possible groups of $k$ object from $n$ objects First of all, I'm sorry if this has already been answered, but I could not find the same exact problem.
So here it is. I want to know how many different groups of $k$ different objects I can make from $n$ objects, and the order of the objects within the groups does not matter.
Lets take an example. Sat, I have 10 objects. How many groups of 2 objects can I make from those 10?
I listed them and I found 45 :
($n1$,$n2$), ($n1$,$n3$),....,($n9$,$n10$)
So as you can see, each number appears in several groups (since it is paired with each other number)
I did the same for $k$ = 3 and found 124, and for $k$ = 4 and found 201. (I may be absolutely wrong however since I listed them all on paper, and may have jumped one or two of them)
However I am unable to find a general definition of this, and I thought maybe you guys could help me finding the function I'm looking for.
The only thing I know so far is that $f(k)$ = $f(n-k)$ (if $k < n$)
So in my example, $f(1)$ = 10 and $f(9)$ = 10 (since it's basically the same question: "how many different groups of $n - k$ objects can I take away from $n$?")
Sorry for my approximative use of english, and thank you for reading.
 A: The number of subsets of a set with $n$ elements is called a combination.  The number of such subsets is 
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
Given a set with $n$ elements, we can select the first element in $n$ ways, the second element in $n - 1$ ways, the third in $n - 2$ ways, and so forth.  In general, we can select the $k$th element in $n - (k - 1) = n - k + 1$ ways.   Hence, the number of ordered selections (permutations) of $k$ elements from a set of $n$ elements is 
$$P(n, k) = n(n - 1)(n - 2) \ldots (n - k + 1)$$
Notice that if we multiply the numerator and denominator of $P(n, k)$ $(n - k)!$, we obtain
$$P(n, k) = \frac{n(n - 1)(n - 2) \ldots (n - k + 1)(n - k)!}{(n - k)!} = \frac{n!}{(n - k)!}$$
However, we could have selected the same $k$ elements in $k!$ orders.  Hence, the number of subsets (unordered selections) of $k$ elements from an $n$ element set is 
$$C(n, k) = \binom{n}{k} = \frac{n!}{k!(n - k)!}$$
The symbol $\binom{n}{k}$ is read ``$n$ choose $k$''.
A: https://en.wikipedia.org/wiki/Binomial_coefficient
The binomial coefficient $\binom{n}{k}$, read aloud as "$n$ choose $k$", counts the number of subsets of size $k$ from a set of size $n$.  Equivalently, the number of ways of selecting $k$ objects from $n$ distinct objects where order of selection doesn't matter.  Further, there are dozens more scenarios where binomial coefficients appear in various other counting problems.
One can (and should) prove that $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ (where here $n!=1\cdot 2\cdot 3\cdots (n-1)\cdot n=\prod\limits_{k=1}^n k$)
One can also define the binomial coefficients recursively noting that


*

*$\binom{0}{n}=0$ for all $n>0$

*$\binom{n}{0}=1$ for all $n\geq 0$

*$\binom{n}{r}=0$ for all $r>n$

*$\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$
