What is $ {n\choose k}$? This is the Binomial theorem:
$$(a+b)^n=\sum_{k=0}^n{n\choose k}a^{n-k}b^k.$$
I do not understand the symbol 
$ {n\choose k}.$ How do I actually compute this? What does this notation mean? Help is appreciated.
 A: The symbol ${n\choose k}$ is read as "$n$ choose $k$." It represents the number of ways to choose $k$ objects from a set of $n$ objects. It has the following formula
$$ {n\choose k}=\frac{n!}{(n-k)!k!}.$$
Here, 
$$ n!=n(n-1)(n-2)\cdots2\cdot1.$$
A: $$\binom{n}{k}= \frac{n!}{k!(n-k)!}$$
It computes the number of ways we can choose $k$ items out of $n$ items.
A: $n \choose k$ - n choose k - how many different ways there are to pick $k$ items from a set of $n$ elements. The explanation starts from permutations, through combinations, finishing with binomial theory. If you are familiar with the formulas and the ideas behind them feel free to skip some steps.
Permutations
A permutation of a set $\mathcal{S}$ is an arrangement of its elements in a specific order. The number of possible permutations is denoted as $P$.
$$
\begin{aligned}
P^n_n &= n(n-1)(n-2)\cdots1 \\\\
      &= n!
\end{aligned}
$$
The idea behind the formula: on the first pick we can choose from $n$ elements - $n$ possible picks. After picking the first element, we are left with $n-1$ elements from which to choose. Following this logic, eventually, we are left with $2$ elements from which to choose and then finally $1$.
It can be observed that we can prematurely end this process and only choose $k$ elements from $\mathcal{S}$:
$$
P^n_k=n(n-1)(n-2)\cdots(n-(k-1))
$$
The idea is the same just that we stop after our $k$-th pick, during which we had $n-(k-1)$ choices as we already picked $k-1$ elements.
Combinations
The base idea behind combinations is that one is picking $k$ elements from the set $\mathcal{S}$ of size $n$ but not putting them in a sequence - meaning that the order is not important. We can view combinations as creating a subset 
$\mathcal{S'}\subseteq\mathcal{S}$, of size $k$, $k\leq n$. The number of possible ways we can create this subset $\mathcal{S'}$ from the set $\mathcal{S}$ is denoted as $C$ and defined as:
$$
\begin{align}
C^n_k & = {n\choose k} \\\\
  &=\frac{n(n-1)(n-2)\cdots(n-(k-1))}{k!} \\\\
  &= \frac{n!}{(n-k)!k!}
\end{align}
$$ 
Observe how the numerator is equivalent to $P^n_k$ and the denominator to $P^k_k$. 
The idea behind the formula: notice that the extra variability in $P^n_k$ comes from the different permutations ($P^k_k$) each of those combinations ($C^n_k$) can make. To put this in mathematical terms:
$$
\begin{align}
P^n_k &= C^n_kP^k_k \\\\
      &\Rightarrow \\\\
C^n_k &= \frac{P^n_k}{P^k_k} \\\\
      &= \frac{n(n-1)(n-2)\cdots(n-(k-1))}{k!} \\\\
      &= \frac{n!}{k!(n-k)!}
\end{align}
$$
Binomial Coefficients
Let's take the example of $(x+y)^n$ and write it as a product of $n$ binomials:
$$
(x+y)^n = (x+y)(x+y)\cdots(x+y)
$$
If we were to expand this, each resulting term in the expansion will be a product of $n$ terms, one term picked from each of the binomials above. We need to focus only on $x$ (one of the terms) - if we choose to pick $x$ $k$ times then we must pick $y$ $n-k$ times. There are $n$ binomials in the expansions and we are making $k$ picks, therefore, the number of resulting terms in the expansion with the $k$ picks of $x$ is $C^n_k$ or $n\choose k$.
A: $\binom{n}{k}:=\frac{n!}{(n-k)!k!}$
With $n!=n\cdot (n-1)\cdot (n-2)\cdot\dotso\cdot 3\cdot 2\cdot 1=\prod_{i=1}^n i$
A: $ \binom{n}{k} = \frac{n!}{(n-k)!k!} $
It is used to calculate the number of ways "k" events can occur in "n" choices. 
A: $${n\choose k}={n!\over k!(n-k)!}$$
A: Number of words of length $n$ over the alphabet $\{a,b\}$ such that $k$ letters are $a$. When you expand the LHS this characterization immediately implies the RHS.
