Probability notation two numbers stacked inside brackets

Assume $m\geq n$ are nonnegative integers. We define $\binom{m}{n}$ as $$ \binom{m}{n} := \frac{m!}{n!\,(m-n)!} = \dfrac{m(m-1)(m-2)\cdots 3\cdot 2\cdot 1}{(n(n-1)\cdots 3\cdot 2\cdot 1)((m-n)(m-n-1)\cdots 3\cdot 2\cdot 1)}, $$ and read $\binom{m}{n}$ as "$m$ choose $n$". As others have mentioned above, this is called the $\textbf{binomial coefficient}$.

Let's go back to the example $\binom{5}{1}$. One could think of this as the number of ways to choose $1$ object in a bag of $5$ objects. If you have a bag with $5$ objects, how many ways are there to pick one item? There are $5$ ways.

That is, $$ \binom{5}{1}=\frac{5!}{1! (5-1)!} = \frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{1\cdot 4\cdot 3\cdot 2\cdot 1} = 5. $$

Now consider the example $\binom{5}{2}$. How many ways are there to pick two items in a bag with $5$ items (without replacement)?

There are $5$ ways to pick the first item (let's call this item $A$) and then there are $4$ ways to pick the second item (let's call this item $B$). Since the ordering does not matter, we divide $5\cdot 4$ by $2$ to obtain: $$ \frac{5\cdot 4}{2} = 10, $$ which is $$ \binom{5}{2} = \frac{5!}{2!\, 3!} = \frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1\cdot 3\cdot 2\cdot 1} = 10. $$

This is the usual notation for a binomial coefficient: $\binom{123}{45}$ is the number of different $45$-element subsets of a set with $123$ elements in total. It can be computed as $$ \binom{123}{45} = \frac{123!}{45!(123-45)!} $$

In some elementary probability/combinatorics texts, the same concept can also be notated something like ${}^{123}C_{45}$ or ${}_{123}C_{45}$ or $C(123,45)$.

It's the 'choose' function. Say you have 8 people and you want to pick (choose) 3 of them to form a team. Then the number of different teams you can create is $8\choose3$.