Math nomenclature

Question

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

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edit and added

so if i had $$ \binom{0}{0} $$ would that be $$ \dfrac{0!}{0!(0-0)!} $$ ...what would that be...zero?

Answer 1

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.

Answer 2

$$\binom{k}{i}\; \text{ is called a $\color{blue}{\bf{binomial\;coefficient}}$, which is read as "k choose i"}$$

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 $$\displaystyle \;\binom{k}{i} = \frac{k!}{i!(k-i)!} = \frac{k\cdot (k-1)\cdots (k - i + 1)}{i\cdot (i-1) \cdots 2 \cdot 1}$$

Answer 3

$k \choose i$ 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.

Answer 4

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.