Number of ways to take $k$ out of $n$ (1 to $n$) numbers and arrange them in order How many ways are there to take $k$ out of $n$ numbers ($1$ to $n$) and arrange them in order?
Example: $n=4$, $k=2$:
$12, 13, 14$
$23, 24$
$34$
I tried to find an answer on SE but all I got were questions about $k$ consecutive out of $n$ numbers. I only need them to be in order. Sorry if that question has been asked already.
Thank you so much
 A: Hint: Given $k$ distinct numbers, there is only one way to order them, so all you need is the number of ways of choosing $k$ numbers from $n$.
A: You have $n$ choices for the first, $n-1$ choices for the second, $n-2$ for the third, on until $n-k+1$ for the last.  Then
$$n(n-1)(n-2)\ldots(n-k+1)=\frac {n!}{(n-k)!}$$
For the edited question, there are $${n \choose k}=\frac {n!}{k!(n-k)!}$$ ways to choose $k$ items out of $n$.  The above version has a factor $k!$ because it can put them in any order.  See Wikipedia on combinations.
A: What you're looking for is called the binomial coefficient. Let's start from the basics, and look at what the binomial coefficient is, in a combinatoric sense:
For $k,n\in\mathbb N$, the binomial coefficient is defined as:
$$\binom{n}{k} :=\left|\binom{\{1,...,n\}}{k}\right|$$
where
$$\binom{\{1,...,n\}}{k} := \{A\in\{1,...,n\}\mid |A|=k\}
$$
In other words, the binomial coefficient precisely counts the amount of $k$-element sets you can make from the set $\{1,...,n\}$.
Now, why is it that the amount of unordered sets equals the amount of sets, if we impose an order on them?
Let's look at one set, e.g. $\{1,2,3\}$. Using this set, we can create the following permutations:
$$
[1,2,3] ; [1,3,2] ; [2,1,3] ; [2,3,1] ; [3,1,2] ; [3,2,1]
$$
Each of these permutations however is nothing but our set, on which we additionally imposed an order. For our first case, $[1,2,3]$, the order would be the regular ascending order - i.e. $1<2<3$, and it's easy to see, that this order is only true for the first case, and for none of the others we listed there (i.e. $2>1$ in $[2,1,3]$).
Generally we have, that if we impose on a set a total order, there is precisely one permutation of this set which fulfills the order.
To finish this in the good, combinatorial fashion we now show that there exists a bijection between the binomial coefficient $\binom{\{1,...,n\}}{2} $ and $ \{(x,y)\in\{1,...,n\}^2\mid x<y\}$, the set of all ordered pairs with two elements.
For this we create the function
$$\phi:\binom{\{1,...,n\}}{2} \to \{(x,y)\in\{1,...,n\}^2\mid x<y\}
\\
\{a,b\}\mapsto (\min(a,b),\max(a,b))
$$
We can show that this function is a bijection in multiple ways. The cleanest is to show that it is injective and surjective. So let's take a look:
The function is surjective, as for any $(a,b)\in \{(x,y)\in\{1,...,n\}^2\mid x<y\}$ we can find a set $A\in \binom{\{1,...,n\}}{2}$, namely the set $\{a,b\}$.
The function is injective, as if we'd have two tuples $\{a_1,b_1\},\{a_2,b_2\}\in \binom{\{1,...,n\}}{2}$ with 
$$
\phi(\{a_1,b_1\})=\phi(\{a_2,b_2\})
$$
we'd know that their minimum and maximum are equal, i.e. 
$$\min(a_1,b_1) =\min(a_2,b_2)
\\ \land\\
\max(a_1,b_1) =\max(a_2,b_2)$$
As we have though that $a_1\neq b_1 \land a_2\neq b_2$ (this follows from the forms of the sets in $\binom{\{1,...,n\}}{2}$), this already means that the both sets $\{a_1,b_1\},\{a_2,b_2\}$ are identical (they have the same lower, and the same higher element, and they have exactly two elements, and the higher and lower element aren't identical).
Therefore $\phi$ is bijective, and thus
$$\left|\{(x,y)\in\{1,...,n\}^2\mid x<y\}\right|= 
\left|\binom{\{1,...,n\}}{2}\right| =
\binom{n}{2}  $$ 
A: This is the number of $k$-combinations out of a set of $n$ elements, then it's up to you to order these $k$ elements, so the number sought for is simply
$$ \binom nk =\frac{n!}{k!(n-k)!}. $$
