diff --git a/docs/1b/ece108.md b/docs/1b/ece108.md
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@@ -734,3 +734,115 @@ Where $x,y,z$ are elements in $X$, and $p,q,r$ are arbitrary proposition results
 
 - Asymmetric relations must be oppositely symmetrical across the main diagonal. The main diagonal also must be false.
 - Antisymmetric relations must be false only if there is a true.
+
+### Transitivity
+
+A **transitive** relation links related terms. For example, $a<b$ and $b<c$ implies $a<c$.
+
+$$\forall x,y,z\in X,\left<x,y\right>\in R\wedge\left<y,z\right>\in R\implies\left<x,z\right>\in R$$
+
+## Orders
+
+!!! definition
+    - A **partial order** is reflextive, antisymmetric, and transitive.
+
+
+A **partially ordered set (poset)** is a set $S$ partially ordered with relation $R$.
+
+$$\left<S,R\right>\text{ on } P=R_{S,P}$$
+
+!!! example
+    $R_{\mathbb Z,\geq}$ is a poset. $\left<\mathcal P(A),\subseteq\right>$ on $A$ is also a poset.
+
+A **strict poset** is irreflexive, asymmetric, and transitive.
+
+A **total order** is a strict poset such that the relation is defined between every possible pair on the set.
+
+$$\forall x,y\in S,xPy\wedge yPx\in\left<S,P\right>$$
+
+### Equivalence relations
+
+An **equivalence class** is a criterion that determines whether two objects are equivalent. The original set must be the union of all equivalence classes.
+
+!!! example
+    The following are all in the equivalence class $=_1$: $\{1,\frac 2 2,\frac 3 3,\frac 4 4,...\right}$
+    
+## Combinatorics
+
+!!! definition
+    - **and** usually requires you to multiply sets together.
+    - **or** usually requires you to add then subtract unions.
+
+The number of ways to choose exactly one element from finite sets is the product of their dimensions.
+
+$$|A_1||A_2|...|A_n|$$
+
+!!! example
+    The number of unique combinations (including order) from four dice is $|6|^4$.
+
+### Ordered with replacement
+
+These problems count order as separate permutations and replace an item after it is taken for the future. If there are $n$ outcomes, and $m$ events that take one of those outcomes:
+
+$$P=n^m$$
+
+To pick $m$ items out of $n$ elements:
+
+$$P(n,m)=\frac{n!}{(n-m!)}$$
+
+If there are duplicates that would otherwise result in an identical string, divide the result by $m!$, where $m$ is the number of repetitions for each duplicate $n_1,n_2,...$.
+
+$${n\choose n_1!n_2!n_k!}=\frac{n!}{n_1!n_2!...n_k!}$$
+
+!!! example
+    The number of permutations of "ECE119" has two characters that have duplicates. Therefore, the number of possibilities is:
+    $$\frac{6!}{2!2!}$$
+
+### Unordered with replacement
+
+To rearrange $n$ unique items, the number of possibilities is:
+
+$$n!$$
+
+To choose $n$ items $m$ times, regardless of order, the number of possibilities is:
+
+$${n\choose m}=\frac{n!}{(n-m)!m!}={n\choose(n-m),m}$$
+
+Clearly ${n\choose m}=0$ if $m>n$ or $m<0$.
+
+To choose $k$ out of $n$ items one time, multichoose can be used:
+
+$$\left({n\choose k}\right)={n+k-1\choose k}={n-1+k\choose n-1,k}$$
+
+### Binomial coefficients
+
+A **slack variable** is used to change inequalities into equalities.
+
+!!! example
+    If solving $x+y\leq 7$, setting $z=7-(x+y)$ to make everything the same domain ($\mathbb Z^+_0$) to use choose.
+
+**Pascal's identity** defines the choose operator recursively.
+
+$${n\choose m}={n-1\choose m-1}+{n-1\choose m}$$
+
+The **binomial theorem** expands a binomial.
+
+$$\forall a,b\in\mathbb R,(a+b)^n=\sum^n_{i=0}{n\choose i}a^{n-i}b^i$$
+
+The sum of choosing integers is its power to 2. Therefore, a finite set with dimension $n$ must have exactly $2^n$ possible subsets.
+
+$$\forall n\in\mathbb Z^+_0,\sum^n_{k=0}{n\choose k}=2^n$$
+
+### Inclusion-exclusion
+
+The inclusion-exclusion principle removes duplicate counting.
+
+$$|A\cup B|=|A|+|B|-|A\cap B|$$
+
+This can be extended to 3+ sets, proven by a bijection to $\mathbb N_{|A| + |B|+|A\cap B|}$:
+
+$$|A\cup B\cup C|=|A| + |B| + |C| - (|A\cap B| + |A\cap C| + |B\cap C|)-|A\cap B\cap C|$$
+
+If $B$ is a subset of $A$, the dimension of $B$ is related to that of $A$.
+
+$$B\subseteq A\implies|B|=|A|-|\overline B|$$