2019 fall midterm and 2019 winter final don't seem to have solutions …

…in there

src/content/questions/comp2804/2019-fall-final/3/solution.md

@@ -15,6 +15,9 @@ Let's break me down
     The remaining 40 positions can be any of the 2 letters: $ 2^{40} $ <br/> 
     Thus, the total number of strings is $ \binom{85}{15} \cdot \binom{70}{30} \cdot 2^{40} $
 </ul>
-Now, we can determine $A \cup B$ 
-$ |A \cup B| = |A| + |B| - |A \cap B| $ 
+
+Now, we can determine $A \cup B$
+
+$ |A \cup B| = |A| + |B| - |A \cap B| $
+
 $ |A \cup B| = \binom{85}{15} \cdot 3^{70} + \binom{85}{30} \cdot 3^{55} - \binom{85}{15} \cdot \binom{70}{30} \cdot 2^{40} $

src/content/questions/comp2804/2019-fall-final/4/solution.md

@@ -14,6 +14,9 @@ Let's do this uwu
     We choose 12 of the 50 cider bottles: $ \binom{50}{12} $ <br/> 
     $|A \cap B| = \binom{20}{12} \cdot \binom{50}{12} $
 </ul>
-Now, we can determine $A \cup B$ 
-$ |A \cup B| = |A| + |B| - |A \cap B| $ 
+
+Now, we can determine $A \cup B$
+
+$ |A \cup B| = |A| + |B| - |A \cap B| $
+
 $ |A \cup B| = \binom{20}{12} \cdot 2^{50} + \binom{50}{12} \cdot 2^{20} - \binom{20}{12} \cdot \binom{50}{12} $

src/content/questions/comp2804/2019-fall-final/5/solution.md

@@ -14,6 +14,9 @@
     We choose any subset of the 50 cider bottles: $ 2^{50} $ <br/> 
     $ |C| = \binom{20}{2} \cdot 2^{50} = 190 \cdot 2^{50} $
 </ul>
-Now, we can determine the number of subsets that contain at least 3 beer bottles 
-$ |S| - |A| - |B| - |C| $ 
+
+Now, we can determine the number of subsets that contain at least 3 beer bottles
+
+$ |S| - |A| - |B| - |C| $
+
 $ = 2^{70} - 2^{50} - 20 \cdot 2^{50} -\binom{20}{2} \cdot 2^{50} $

src/content/questions/comp2804/2019-fall-final/7/solution.md

@@ -12,6 +12,9 @@ Assuming you have a sea of bananas, let's put dividers to split them off into di
     <li> $x_6$ is the number of bananas between the fifth and sixth divider
     <li> $x_7$ is the number of bananas to the right of the sixth divider
 </ul>
-Now, we have 6 dividers and 25 bananas, so we have 31 objects in total 
+
+Now, we have 6 dividers and 25 bananas, so we have 31 objects in total
+
 Since each object has a position, we can just move the dividers around to change any x.
+
 We choose 6 spots out of the 31 to be dividers: $ \binom{31}{6} $

src/content/questions/comp2804/2019-winter-midterm/11/solution.md

@@ -1,13 +1,7 @@
-textbf{Plaintext Question:} The function $ f: \mathbb{N} rightarrow \mathbb{N} $
-is recursively defined as follows:
-$ f(0) = 6, $
-$ f(n) = 4 \cdot f(n-1) + 2^{n} \text{ if } n geq 1. $
-Which of the following is true for all integers $ n geq 0 $? 
-textbf{Correct Answer: } B 
-textbf{Relevant Theorems: } Recursive Functions (4.1) 
-textbf{Explanation: } 
-We can find the first few values of $ f(n) $ to find a pattern: 
+We can find the first few values of $ f(n) $ to find a pattern:
+
 $ f(1) = 4 \cdot f(0) + 2^1 = 4 \cdot 6 + 2 = 26 $
+
 $ f(2) = 4 \cdot f(1) + 2^2 = 4 \cdot 26 + 4 = 108 $
 
 <ul>
@@ -18,4 +12,5 @@ $ f(2) = 4 \cdot f(1) + 2^2 = 4 \cdot 26 + 4 = 108 $
     <li> $ f(n) = 8 \cdot 4^{n} - 2^{n+1} $ <br/> 
     $ f(1) = 8 \cdot 4^{1} - 2^{2} = 32 - 4 = 28 $
 </ul>
+
 Thus, the correct answer is $ f(n) = 7 \cdot 4^{n} - 2^{n} $.