Math nomenclature I kinda feel ashamed about asking this, but could someone explain me what this means?
$$
\binom{k}{i}
$$
That should indicate the number of nodes at a certain depth in a binomial heap..but I can't remember what that actually means and I wasn't successful with Google!
Any explanation/link/idea would be awesome!
Thanks
==================================
edit and added
so if i had 
$$
\binom{0}{0} 
$$
would that be 
$$
\dfrac{0!}{0!(0-0)!}
$$
...what would that be...zero?
 A: I assume you mean $$\dbinom{n}k$$
$\dbinom{n}k$ is a short hand for the number of ways in which you can choose $k$ objects from $n$ distinguishable objects. These number are called the binomial coefficients. Some textbooks and articles also denote this as $C(n,k)$.
As an example, if we have three colored balls, $\color{red}{\text{red}}$, $\color{blue}{\text{blue}}$ and $\color{brown}{\text{brown}}$, then there are three ways of choosing two balls.
\begin{matrix}
\color{red}{\text{red}} & \color{blue}{\text{blue}}\\
\color{blue}{\text{blue}} & \color{brown}{\text{brown}}\\
\color{brown}{\text{brown}} & \color{red}{\text{red}}
\end{matrix}
Note that when we say we choose, we are not interested in the order in which these are picked i.e. $\color{red}{\text{red}} \,\, \color{blue}{\text{blue}}$ and $\color{blue}{\text{blue}} \,\, \color{red}{\text{red}}$ refer to the same choice of two balls. Hence, we have $\dbinom{3}2 = 3$.
Similarly, if we have $5$ colored balls, say $\color{red}{\text{red}}$, $\color{blue}{\text{blue}}$, $\color{brown}{\text{brown}}$, $\color{orange}{\text{orange}}$, and $\color{lightgreen}{\text{green}}$, there are $10$ ways of choosing $3$ balls.
\begin{matrix}
\color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{brown}{\text{brown}}\\
\color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{orange}{\text{orange}}\\
\color{red}{\text{red}} & \color{blue}{\text{blue}} & \color{lightgreen}{\text{green}}\\
\color{red}{\text{red}} & \color{orange}{\text{orange}} & \color{brown}{\text{brown}}\\
\color{red}{\text{red}} & \color{orange}{\text{orange}} & \color{lightgreen}{\text{green}}\\
\color{red}{\text{red}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}}\\
\color{blue}{\text{blue}} & \color{brown}{\text{brown}} & \color{orange}{\text{orange}}\\
\color{blue}{\text{blue}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}}\\
\color{blue}{\text{blue}} & \color{orange}{\text{orange}} & \color{lightgreen}{\text{green}}\\
\color{orange}{\text{orange}} & \color{brown}{\text{brown}} & \color{lightgreen}{\text{green}}
\end{matrix}
In general,
$$\dbinom{n}r = \dfrac{n!}{r!(n-r)!}$$
where $k! = k \times (k-1) \times (k-2) \times \cdots \times 2 \times 1$.
The name binomial coefficient arises from binomial theorem. When we expand $(x+y)^n$, the coefficient of $x^k y^{n-k}$ is given by $\dbinom{n}k$ i.e.
$$(x+y)^n = \sum_{k=0}^n \dbinom{n}k x^k y^{n-k}$$
There are a lot of wonderful properties these binomial coefficients satisfy and I highly recommend you to go through the wiki-page for these properties.
A: $$\binom{k}{i}\; \text{ is called a $\color{blue}{\bf{binomial\;coefficient}}$, which is read as "k choose i"}$$
See Binomial Coefficient.
"k choose i" comes from the fact that if gives you the number of ways to choose $i$ elements from a set of $k$ elements.
The term "binomial coefficient" makes explicit its relationship to the binomial theorem. When we expand $(x+y)^n$, the coefficient of $x^k y^{n-k}$ is given by $\large\binom{n}k$ i.e.
$$(x+y)^n = \sum_{k=0}^n \dbinom{n}k x^k y^{n-k}$$
You can compute your binomial coeffient by noting that $$\displaystyle \;\binom{k}{i} = \frac{k!}{i!(k-i)!} = \frac{k\cdot (k-1)\cdots (k - i + 1)}{i\cdot (i-1) \cdots 2 \cdot 1}$$
A: $k \choose i$ is a binomial coefficient.
It is the coefficient of $x^i$ in expanding $(1+x)^k$.
It is also the number of $i$-element subsets of a $k$-element set.
It can be computed via $\frac{k!}{i!(k-i)!}$ or with several recursive formulas.
Values are often listed with help of Pascal's triangle.
A: For the question in the edit: We have $0! = 1$, so $\binom{0}{0} = 1$ which is the first entry in Pascal's triangle. 
A previous question on this site asked about the definition of zero factorial; there are several excellent answers there. As Zhen Lin points out in a comment, the best explanation of why $0! = 1$ depends on how you define the factorial.
