We define the binomial coefficient of choose as $$ {r\choose k} := \begin{cases}\dfrac{r^\underline k}{k!} & k \in \Bbb N \
0 & \text{otherwise},\end{cases}$$where is the th falling factorial. This definition is extremely useful since, it lets us choose from more than just natural numbers, and let .
Some notable examples are $${r\choose 0} = 1, \quad {r\choose 1} = r, \quad {r\choose 2} = \frac{r(r-1)}{2}.$$
If we let , then we have the following formula $${n \choose k} = \frac{n!}{k!(n-k)!}.$$This leads to a symmetric formula for the binomial coefficients when $${n \choose k} = {n \choose n-k}.$$By the same definition, we get a general identities $${r\choose k} = \frac{r}{k} {r-1\choose k-1}, \qquad k {r\choose k} = r{r-1\choose k-1}, \qquad (r-k){r\choose k} = r{r-1\choose k}. $$We also get the addition formula, or the recursive formula from this identities $${r\choose k} = {r-1 \choose k} + {r-1\choose k-1}.$$If we inductively apply this formula we get the a general formula: $$\sum_{k \le n} {r+k \choose k} = {r+n+1\choose n}.$$We also have another identity for $$\sum_{k = 0}^n {k\choose m } = {n+1\choose m+1}.$$This leads to a nice special case where , $$\sum_{k =0}^n k =\sum_{k =0}^n {k \choose 1} = {n+1\choose 2} = \frac{n(n+1)}{2}.$$The general result is equivalent to the result of Discrete Calculus $$\sum_{k=0}^n k^{\underline m} = \frac{(n+1)^{\underline{m+1}}}{m+1},$$which I found amusing. We can also see that $$\nabla \left({x\choose m}\right) = {x\choose m-1}, \qquad \nabla \left({x\choose m}\right) = {x-1\choose m-1}.$$
We can prove a inductively on , that $$\sum_{k = 0}^m {r\choose k}\left(\frac{r}{2}- k\right) = \frac{m+1}{2} {r\choose m+1}. $$
We get the binomial theorem and the extended binomial theorem. For and then $$(x+y)^n = \sum_{k = 0}^n {n \choose k} x^k y^{n-k},$$if with and , then $$(x+y)^r = \sum_{k = 0}^\infty {r \choose k}x^k y^{n-k}. $$
If we look for the connection to the negative binomial coefficients then we get that $${-r \choose k } =(-1)^k{r+k-1\choose k}.$$This has a nice consequence of $$\sum_{k \le m} {r\choose k}(-1)^k = (-1)^m{r-1\choose m}.$$
If we ask about the partial sums of the binomial coefficients, we know it doesn't have a closed form, but we can have a certain equality $$\sum_{k \le m} {m +r \choose k} x^k y^{n-k} = \sum_{k \le m} {-r\choose k} (-x)^k(x+y)^{m+k}, \quad \text{for }. $$Let us not that this is a more general version of the identity above, since if we let and , then we get the original identity. Now, if and , then $$2^{2m}= \sum_{k \le m} {2m+1\choose k} = \sum_{k \le m} {m+k \choose k} 2^{m-k}.$$thus $$2^m = \sum_{k \le m}{m+k\choose k} 2^{-k}.$$
Let and , then $${r\choose m}{m \choose k} = {r\choose k}{r-k\choose m-k}.$$
We have a nice generalisation of the binomial coefficients into the multinomial coefficients then $$ {a_1 + a_2 + \dots + a_n \choose a_1, a_2, \dots, a_n} :=\frac{(a_1+a_2 +\dots +a_n)!}{a_1!a_2! \cdots a_n !} = {a_1 +a_2 + \dots +a_n\choose a_2 +\dots+ a_n} \cdots {a_{n-1}+a_n\choose a_n}.$$
With these tools we get the the following generalisation of the binomial theorem, called the multinomial theorem: $$\left(\sum_{k = 1}^n x_i\right)^m = \sum_{|\alpha | = m} {m \choose \alpha} x^\alpha$$where is a multi-index, and .
The following identity is called Vandermonde's Convolution: $$\sum_{k = 0}^n {r\choose k} {s\choose n-k} ={r+s\choose n}\quad \text{for }. $$From this we get a couple of identities $$\begin{align*}
\sum_{k} {r\choose m+k} {s\choose n-k} &={r+s\choose m+n}\qquad \text{for }. \ \
\sum_{k} {\ell \choose m+k}{s\choose n+k} &= {\ell +s \choose \ell -m+n}
\qquad \text{for and }.\ \
\sum_{k} {\ell \choose m+k}{s+k\choose n}(-1)^k &= (-1)^{m+1}{s-m \choose n-\ell} \qquad \text{for and }.\ \
\sum_{k \le \ell} {\ell-k \choose m}{s \choose k-n} (-1)^k &= (-1)^{1+m} {s-m-\ell \choose \ell -m -n} \qquad \text{for }. \ \
\sum_{k = 0}^\infty {\ell -k \choose m}{q+k\choose n} &= {\ell +q +1\choose m+n+1}\qquad \text{for for }.
\end