Prove the binomial identity

Question

Show that for $k = 0, 1, 2, \ldots , n$, and for $n = 1, 2, 3, \ldots$ we have: $$ {n \choose k} = {{n-1} \choose {k-1}} + {{n-1} \choose k}$$

I know that there are a lot of solutions for the Newton's equal but I'm really sure what to answer.

Answer 1

$$LHS={n\choose k}=\frac{n!}{(n-k)!k!}$$ $$RHS={n-1\choose k-1}+{n-1\choose k}=\frac{(n-1)!}{(n-k)!(k-1)!}+\frac{(n-1)!}{(n-k-1)!k!}=\frac{(n-1)!}{(n-k-1)!(k-1)!}\left[\frac{1}{n-k}+\frac1k\right]=\frac{(n-1)!}{(n-k-1)!(k-1)!}\left[\frac{n}{(n-k)k}\right]=\frac{n!}{(n-k)!k!}=LHS$$

Answer 2

You have a total of $n$ objects. You have two possible techniques to choose a $k$-set:

Method 1: Choose $k$ elements from the $n$ objects directly. This gives the count on the LHS.

Method 2: Single out an arbitrary element $a$. Now there are the $k$-sets containing $a$ (by taking $k-1$ of the remaining $n-1$ elements), and there are the $k$-sets not containing $a$ (by taking $k$ of the remaining $n-1$ elements). This gives the count on the RHS.

Answer 3

The formula also reflects \begin{align*} (1+x)^n=(1+x)^{n-1}(1+x) \end{align*}

Indeed, using $[x^n]$ to denote the coefficient of $x^n$ in an expression we obtain \begin{align*} \binom{n}{k}&=[x^k](1+x)^n\\ &=[x^k](1+x)^{n-1}(1+x)\\ &=[x^k](1+x)^{n-1}+[x^k]x(1+x)^{n-1}\\ &=\binom{n-1}{k}+\binom{n-1}{k-1} \end{align*}

Here we use $[x^k]xA(x)=[x^{k-1}]A(x)$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

An interesting one is given by the $\ds{Binomial\ Integral\ Representation}$ $$ {a \choose b} = \oint_{\verts{z} = R}{\pars{1 + z}^{a} \over z^{b + 1}} \,{\dd z \over 2\pi\ic}\,,\qquad b \in \mathbb{Z} $$ $\ds{R > 0}$ is chosen such that the $\ds{\pars{1 + z}^{a}}$-expansion, in powers of $\ds{z}$, converges.

Namely, \begin{align} {n \choose k} & = \oint_{\verts{z} = R}{\pars{1 + z}^{n} \over z^{k + 1}}\,{\dd z \over 2\pi\ic} = \oint_{\verts{z} = R}{\pars{1 + z}^{n - 1}\pars{1 + z} \over z^{k + 1}}\,{\dd z \over 2\pi\ic} \\[5mm] & = \oint_{\verts{z} = R}{\pars{1 + z}^{n - 1} \over z^{k + 1}}\,{\dd z \over 2\pi\ic} + \oint_{\verts{z} = R}{\pars{1 + z}^{n - 1} \over z^{k}}\,{\dd z \over 2\pi\ic} = \bbx{\ds{{n - 1 \choose k} + {n - 1 \choose k - 1}}}\,,\qquad \end{align}