From Wikipedia's webpage on "binomial coefficients":
"The symbol $\tbinom {n}{k}$ is usually read as "$n$ choose $k$" because there are $\tbinom {n}{k}$ ways to choose an (unordered) subset of $k$ elements from a fixed set of $n$ elements.
Arranging the numbers $\tbinom {n}{0},\tbinom {n}{1},\ldots ,\tbinom {n}{n}$ in successive rows for $n=0,1,2,\ldots$ gives a triangular array called Pascal's triangle, satisfying the recurrence relation
$$\begin{align}
\binom {n}{k} & ={\binom {n-1}{k}}+{\binom {n-1}{k-1}}. \\
\end{align}$$
Commonly, a binomial coefficient is indexed by a pair of integers $n ≥ k ≥ 0$ and is written $\tbinom {n}{k}$. It is the coefficient of the $x^k$ term in the polynomial expansion of the binomial power $(1 + x)^n$, and it is given by the formula
$$\begin{align}
\binom {n}{k} & ={\frac {n!}{k!(n-k)!}}. \qquad\qquad
\end{align}$$
For example, the fourth power of $1 + x$ is
$$\begin{aligned}(1+x)^{4}&={\tbinom {4}{0}}x^{0}+{\tbinom {4}{1}}x^{1}+{\tbinom {4}{2}}x^{2}+{\tbinom {4}{3}}x^{3}+{\tbinom {4}{4}}x^{4}\\&=1+4x+6x^{2}+4x^{3}+x^{4},\end{aligned}$$
and the binomial coefficient $\tbinom {4}{2} ={\tfrac {4!}{2!2!}}=6$ is the coefficient of the $x^2$ term.
The section titled: "computing the value of binomial coefficients" explains:
Recursive formula
One method uses the recursive, purely additive, formula
$$\binom {n}{k} = \binom {n-1}{k-1} + \binom {n-1}{k} \quad \text{for all integers } n,k:1\leq k\leq n-1, \qquad $$
with initial/boundary values
$${\binom {n}{0}}={\binom {n}{n}}=1\quad {\text{for all integers }}n\geq 0, \qquad \qquad \qquad \quad \qquad$$
Multiplicative formula
A more efficient method to compute individual binomial coefficients is given by the formula
$$\binom {n}{k} ={\frac {n^{\underline {k}}}{k!}}={\frac {n(n-1)(n-2)\cdots (n-(k-1))}{k(k-1)(k-2)\cdots 1}}=\prod _{i=1}^{k}{\frac {n+1-i}{i}},\qquad $$
where the numerator of the first fraction $n^{\underline {k}}$ is expressed as a falling factorial power. This formula is easiest to understand for the combinatorial interpretation of binomial coefficients. The numerator gives the number of ways to select a sequence of $k$ distinct objects, retaining the order of selection, from a set of $n$ objects. The denominator counts the number of distinct sequences that define the same $k$-combination when order is disregarded.
Due to the symmetry of the binomial coefficient with regard to $k$ and $n−k$, calculation may be optimised by setting the upper limit of the product above to the smaller of $k$ and $n−k$.
Factorial formula
Finally, though computationally unsuitable, there is the compact form, often used in proofs and derivations, which makes repeated use of the familiar factorial function:
$${\binom {n}{k}}={\frac {n!}{k!\,(n-k)!}}\quad {\text{for }}\ 0\leq k\leq n, \qquad \qquad \qquad \qquad \qquad \qquad \qquad $$
where $n!$ denotes the factorial of $n$. This formula follows from the multiplicative formula above by multiplying numerator and denominator by $(n − k)!$; as a consequence it involves many factors common to numerator and denominator. It is less practical for explicit computation (in the case that $k$ is small and $n$ is large) unless common factors are first cancelled (in particular since factorial values grow very rapidly).
See also: "Injective functions from $\underline{N}$ to $\underline{X}$, up to a permutation of $\underline{N}$" and the "Twelvefold Way", or confirm this at the MathForum.
What is the algorithm for counting combinations?
// pseudo code
start count_combinations( k , n ) {
if (k = n) return 1;
if (k > n/2) k = n-k;
res = n-k+1;
for i = 2 by 1 while i < = k
res = res * (n-k+i)/i;
end for
return res;
end
So for Catalan numbers, a sequence of positive integers, where the $n$th term in the sequence denoted $C_n$, is found in the following formula:
$$C_n = (2n)! / ((n + 1)!n!)$$
The $n$ factorial is equal to the product of all of the integers from $n$ down to $1$.
$$(n) ⋅ (n - 1) ⋅ (n - 2) ⋅ … ⋅ 2 ⋅ 1$$
Without using factorial that's:
$$C_{n}={2n \choose n}-{2n \choose n+1}={1 \over n+1}{2n \choose n}\quad {\text{ for }}n\geq 0,$$
and
$$C_{0}=1\quad {\text{and}}\quad C_{n+1}={\frac {2(2n+1)}{n+2}}C_{n}.$$