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+---
+title: Fundamental Counting
+author: Alexis_Fx & Dylancode
+---
+## Tree Diagram
+### What is tree diagram
+As things get more complex, we are required an effective tool to organise and visualise objects. Diagrams satisfy both requirement. Consequently, Diagrams are delevloped and used as one way for counting. One of them is tree diagram, an early and well-used one due to its simplicity.
+``` mermaid
+flowchart LR
+    S1["Sub Event 1"] --> A["Element 1"] & B["Element 2"] & C["Element 3"]
+    S2(["Sub Event 2"]) --> A1(["Element 4"]) & B2(["Element 5"]) & C3(["Element 6"]) & D4(["Element 7"])
+    n1["Main Event"] --> S1 & S2
+
+    S1@{ shape: rounded}
+    A@{ shape: rounded}
+    B@{ shape: rounded}
+    C@{ shape: rounded}
+    n1@{ shape: diam}
+```
+
+### Example
+```mermaid
+flowchart TD
+    A[Geometry] --> B(Circles)
+    A[Geometry] --> C(Polygons)
+    A[Geometry] --> D(Lines)
+    B --> E(Relative Positions)
+    B --> F(Equations)
+    B --> P(Theorems)
+    C --> G(Triangles)
+    C --> J(Rectangles)
+    D --> I(Relative Positions)
+    D --> K(Slopes)
+    D --> L(Equations)
+```
+## Addition Principle 
+### Statement  
+If one event can occur in $m$ ways and a second event with no common outcomes can occur in $n$ ways, then the first or second event can occur in $m+n$ ways.
+### Example 1 - Addition Priciple
+#### Problem
+In how many ways can we get a drink from a menu with 5 soda options and 8 boba options.
+#### Solution
+Event "we get a soda" can occur in 5 ways and event "we get a boba" can occur in 8 ways. If we get a drink, it has to be from the menu ,or more precisely, either from the soda options or the boba options. Since these two events contribute to form a bigger event which is "we get a drink", according to addition principle, total ways for event "we get a drink" to occur are $5+8=13$ ways 
+```mermaid
+flowchart LR
+    S(["Soda"]) --> A(["Soda no.1"]) & B(["Soda no.2"]) & C(["Soda no.3"]) & D(["Soda no.4"]) & E(["Soda no.5"])
+    Bo(["Boba"]) --> A1(["Boba no.1"]) & B2(["Boba no.2"]) & C3(["Boba no.3"]) & D4(["Boba no.4"]) & E5(["Boba no.5"]) & F5(["Boba no.6"]) & H5(["Boba no.7"]) & K5(["Boba no.8"])
+    n1["Drink_Menu"] --> S & Bo
+
+    n1@{ shape: diam}
+```
+Making a tree diagram like figure above, you can easily find how many outcomes there are just by normal counting.
+
+## Multiplication Principle
+### Statement
+If one event can occur in $m$ ways and a second event can occur in $n$ ways __*after the first event has occurred*__, then the two events can occur in $m \cdot n$ ways.
+```mermaid
+flowchart LR
+    S1["Sub Event 1"] --> S2(["Sub Event 2"])
+    S2 --> n2["Sub Event 3"]
+    n1["Main Event"] --> S1
+
+    S1@{ shape: rounded}
+    n1@{ shape: diam}
+```
+Each Sub Event contains elements from its set. In fact, if we "unbox" a subset , it would look like this (the figure below show subset/subevent 1 with 4 elements)
+```mermaid
+flowchart LR
+    S1["Sub_1 Element 1"] --> S2(["Sub Event 2"])
+    S2 --> n2["Sub Event 3"]
+    n1["Main Event"] --> S1 & n3["Sub_1 Element 2"] & n4["Sub_1 Element 3"] & n5["Sub_1 Element 4"]
+    n6(["Sub Event 2"]) --> n7["Sub Event 3"]
+    n8(["Sub Event 2"]) --> n9["Sub Event 3"]
+    n10(["Sub Event 2"]) --> n11["Sub Event 3"]
+    n3 --> n6
+    n4 --> n8
+    n5 --> n10
+
+    S1@{ shape: rounded}
+    n1@{ shape: diam}
+    n3@{ shape: rounded}
+    n4@{ shape: rounded}
+    n5@{ shape: rounded}
+    n7@{ shape: rect}
+    n9@{ shape: rect}
+    n11@{ shape: rect}
+```
+"Unbox" both subevent 1 and 2
+
+```mermaid
+flowchart LR
+    S1["Sub_1 Element 1"] --> n12(["Sub_2 Element 1"]) & n13(["Sub_2 Element 2"]) & n14(["Sub_2 Element 3"])
+    n1["Main Event"] --> S1 & n3["Sub_1 Element 2"] & n4["Sub_1 Element 3"] & n5["Sub_1 Element 4"]
+    n3 --> n15(["Sub_2 Element 1"]) & n16(["Sub_2 Element 2"]) & n17(["Sub_2 Element 3"])
+    n4 --> n18(["Sub_2 Element 1"]) & n19(["Sub_2 Element 2"]) & n20(["Sub_2 Element 3"])
+    n5 --> n21(["Sub_2 Element 1"]) & n22(["Sub_2 Element 2"]) & n23(["Sub_2 Element 3"])
+    n12 --> n11["Sub Event 3"]
+    n13 --> n9["Sub Event 3"]
+    n14 --> n7["Sub Event 3"]
+    n15 --> n24["Sub Event 3"]
+    n16 --> n25["Sub Event 3"]
+    n17 --> n26["Sub Event 3"]
+    n18 --> n27["Sub Event 3"]
+    n19 --> n28["Sub Event 3"]
+    n20 --> n29["Sub Event 3"]
+    n21 --> n30["Sub Event 3"]
+    n22 --> n31["Sub Event 3"]
+    n23 --> n32["Sub Event 3"]
+
+    S1@{ shape: rounded}
+    n1@{ shape: diam}
+    n3@{ shape: rounded}
+    n4@{ shape: rounded}
+    n5@{ shape: rounded}
+    n11@{ shape: rect}
+    n9@{ shape: rect}
+    n7@{ shape: rect}
+    n24@{ shape: rect}
+    n25@{ shape: rect}
+    n26@{ shape: rect}
+    n27@{ shape: rect}
+    n28@{ shape: rect}
+    n29@{ shape: rect}
+    n30@{ shape: rect}
+    n31@{ shape: rect}
+    n32@{ shape: rect}
+```
+### Explanation
+As you see in the figure above, each element in Sub event 1 can be paired with 1 of the elements in Sub event 2. If there are n elements (or n ways in counting) in Sub event 2, you can pair it with one of the elements in Sub event 1, which give n pairs. And since there are m elements in Sub event 1, there are m of n pairs similarly. Hence prove that the total pairs are $m \cdot n$ pair (or $m \cdot n$ ways in counting).
+### Example 1 - Multiplication Principle
+#### Problem
+In how many ways can we have a lunch consist of a drink from a menu with 5 soda options and 8 boba options, and a dish from the same menu with 3 vegan options and 7 non-vegan options.
+#### Solution
+In this problem, event "having lunch" consist of us choosing a drink and a dish, to be exact, choosing a drink *then* a dish. Consequently, such event is 2 Sub events happen in order : "get a drink" and then "get a dish". According to the principle, total ways we can have a lunch are *(number of ways to get a drink)* $\cdot$ *(number of ways to get a dish)*. We can find them using Addition Principle we established earlier. Number of ways to get a drink are $5+8=13$, number of ways to get a dish are $3+7=10$. Hence the total ways to get a lunch are $13 \cdots 10=130$ ways. 
+