You can add to get to any Number
Let $$a, b \in \mathbb{ Z }$$, then there exists some $$k \in \mathbb{ Z }$$ such that $$a + k = b$$
If we want the equation to hold true, we require some $$k \in \mathbb{ Z }$$ such that $$k = b - a$$, since the integers are closed under subtraction, this value works as verfied by the fact that $$a + \left( b - a \right) = b$$
Numbers from Zero to K
Let $$k \in \mathbb{ N } _ 0$$ and define the set $$X _ k := \left\{ x \in \mathbb{ N } _ 0 : 0 \le x \le k \right\}$$, then $$|X _ k| = k + 1$$

We prove the claim by induction, for the base case if $$k = 0$$, then $$X _ k = \left\{ 0 \right\}$$ thus $$|X _ k| = 1 = 0 + 1 = k + 1$$ as needed.

Let $$n \in \mathbb{ N } _ 1$$, and assume the claim holds for this value, now we want to prove it holds true for $$n + 1$$. We can see that $$X _ (n + 1) = \left\{ x \in \mathbb{ N } ^ 0 : 0 \le x \le n + 1 \right\} = X _ n \cup \left\{ n + 1 \right\}$$, thus $$|X _ (n + 1)| = |X_n| + 1 = n + 1$$ as needed.

The Number of Integers between Two Integers Inclusive
Let $$a, b \in \mathbb{ Z }$$ such that $$a \le b$$ , then $$|\left\{ x \in \mathbb{ Z } : a \le x \le b \right\}| = b - a + 1$$

We know that there is some $$k \in \mathbb{ Z }$$ such that $$b = a + k$$, therefore we're looking at the set $$\left\{ x \in \mathbb{ Z } : a \le x \le a + k \right\}$$, this set is in clear bijection with the set $$\left\{ x \in \mathbb{ Z } : 0 \le x \le k \right\}$$ through the function $$f \left( x \right) = x - a$$, therefore our original set has $$k + 1$$, elements. Moreover $$k = b - a$$ which completes the proof.

Pigeon Hole Principle
Suppose that $$f : X \to Y$$ is between two finite sets and that $$|X| > |Y|$$, then there exists some $$y \in Y$$ and $$a \neq b \in X$$ such that $$f \left( a \right) = f \left( b \right) = y$$
Catalan Numbers
Let $$n \in \mathbb{ N } _ 0$$, then the number of lattice paths from $$\left( 0, 0 \right)$$ to $$\left( n, n \right)$$ which never go above then line $$y = x$$ is said to be the $$n$$-th Catalan Number
Formula For the $$n$$-th Catalan Number
$C \left( n \right) = \frac{1}{n + 1} \binom{2n}{n}$
Bracket String
A bracket string is a finite tuple $$\left( a _ 1, \ldots a _ k \right)$$ such that $$a _ i \in \left\{ \mathtt{(}, \mathtt{)} \right\}$$
For example "())))(" is a bracket string.
Valid Bracket String
A bracket string $$\left( a _ 1, \ldots , a _ k \right)$$ is said to be valid by considering the mapping $$f : \left\{ \mathtt{(}, \mathtt{)} \right\} \to \left\{ -1, 1 \right\}$$ such that for any $$j \in \left[ 1, \ldots k \right]$$ $\sum _ { i = 1 } ^ j f \left( a _ i \right) \le 0 \text{ and } \sum _ { i = 1 } ^ k f \left( a _ i \right) = 0$
Bijection Between Valid Bracket Strings and Lattice Paths
For every valid bracket string involving $$n$$ pairs of brackets, this corresponds to exactly one lattice path from $$\left( 0, 0 \right)$$ to $$\left( n, n \right)$$ which never crosses the diagonal.
Consider the construction of a lattice path $$l$$ from a valid bracket string
• If $$a _ i = \mathtt{(}$$ then $$l _ i = R$$
• If $$a _ i = \mathtt{)}$$ then $$l _ i = U$$
We can prove that this is a lattice path not crossing the diagonal by the definition of valid bracket string.
Number of Valid Bracketings is Equal to the $$n$$-th Catalan Number
The number of distinct arrangement of $$n$$ pairs of left-right parenthesis which close is equal to $$C \left( n \right)$$
This is true because we constructed a bijection valid bracket arrangments and lattice paths, the latter of which we know has size equal to $$C \left( n \right)$$ as needed.
Counting Stars
Let $$S$$ be a nonempty set, and $$*$$ a binary operation on $$S$$, called a star. If $$*$$ is not known to be associative, the expression "the star of $$a, b, c$$ " (for some $$a, b, c \in S$$ ) is ambiguous. It could have one of 12 distinct interpretations: $\begin{array}{llll} (a * b) * c & a *(b * c) & (a * c) * b & a *(c * b) \\ (b * a) * c & b *(a * c) & (b * c) * a & b *(c * a) \\ (c * a) * b & c *(a * b) & (c * b) * a & c *(b * a) . \end{array}$ For every $$n \in \mathbb{Z}^{+}$$, let $$M(n)$$ denote the number of distinct interpretations of the expression "the star of $$a_1, a_2, \ldots, a_n$$ " (where $$a_1, a_2, \ldots, a_n$$ are arbitrary elements in $$S$$ ). Show that $M(n)=\frac{(2 \left( n - 1 \right) )!}{\left( n - 1 \right) !}$
One can count $$M \left( n \right)$$ by first fixing a valid bracket arrangement and then permuting elements with that valid bracket arrangement, moreover note that the star of $$a _ 1, \ldots a _ n$$ contains $$n - 2$$ bracket pairs.