TITLE: Derivation of binomial coefficient in binomial theorem.
QUESTION [15 upvotes]: How was the binomial coefficient of the binomial theorem derived? 
$$\frac{n!}{k!(n-k)!}$$

REPLY [3 votes]: I don't know how the result was originally derived, but here is one calculation.
Let $f(x) = (1+x)^n$ and let us try and find the Maclaurin series for $f(x)$.

The $0$-th derivative $f^{(0)}(x)$ is just $f(x) = (1 + x)^n$ itself.  The first derivative
is $f^{(1)}(x) = n(1+x)^{n-1}$, the second is $f^{(2)}(x) = n(n-1)(1+x)^{n-2}$, and so on.  The
$k$-th derivative is $f^{(k)}(x) =  n(n-1)\cdots (n-k+1)(1+x)^{n-k}$, and finally the $n$-th
derivative is $f^{(n)}(x) = n(n-1)\cdots 2 \cdot 1 = n!$ which is a constant.  Hence, the first
$n+1$ terms of the Maclaurin series for $f(x)$ are
$$f(x) = (1 + x)^n \approx \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k = \sum_{k=0}^n \frac{n(n-1)\cdots (n-k+1)}{k!} x^k.$$
In fact, since  $f^{(n)}(x)$ is a constant, $f^{(k)}(x) = 0$ for all $k > n$.  Thus, the above
is the complete Maclaurin series for $f(x)$; there is no approximation.
The Maclaurin series for $(1+x)^n$ is called the binomial theorem
expansion of $(1+x)^n$, and  the coefficient of $x^k$ in the series
called a binomial coefficient. We denote this coefficient by
$\binom{n}{k}$ and write
$$f(x) = (1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$
Note that 
upon multiplying $\binom{n}{k}$ by $\frac{(n-k)!}{(n-k)!} = 1$ that
$$\binom{n}{k}\times \frac{(n-k)!}{(n-k)!} =
\frac{n(n-1)\cdots (n-k+1)}{k!}\times \frac{(n-k)!}{(n-k)!}
= \frac{n!}{k!(n-k)!}.$$