formatted 2015 winter midterm

src/content/questions/comp2804/2015-winter-midterm/1/solution.md

@@ -1,3 +1,5 @@
 First, we choose 7 of the n students.
+
 Then, we choose one of the remaining n-7 students to be the President.
+
 Thus, the number of ways to choose the Board of Directors is $\binom{n}{7} \cdot (n-7)$.

src/content/questions/comp2804/2015-winter-midterm/10/solution.md

@@ -1,15 +1,18 @@
 We can first calculate values of f$(1)$ and f$(2)$ to see if we can find a pattern.
+
 $f(1) = f(0) + 6(1) - 4 = 15 + 6 - 4 = 17$
+
 $f(2) = f(1) + 6(2) - 4 = 17 + 12 - 4 = 25$
 
 <ul>
-<li> $f(n) = 3n^{2} + 2n + 15$ <br/> 
-$f(1) = 3{(1)}^{2} + 2(1) + 15 = 3 + 2 + 15 = 20$
-<li> $f(n) = 3n^{2} - 2n + 15$ <br/> 
-$f(1) = 3{(1)}^{2} - 2(1) + 15 = 3 - 2 + 15 = 16$
-<li> $f(n) = 3n^{2} + n + 15$ <br/> 
-$f(1) = 3{(1)}^{2} + 1 + 15 = 3 + 1 + 15 = 19$
-<li> $f(n) = 3n^{2} - n + 15$ <br/> 
-$f(1) = 3{(1)}^{2} - 1 + 15 = 3 - 1 + 15 = 17$
+    <li> $f(n) = 3n^{2} + 2n + 15$ <br/> 
+    $f(1) = 3{(1)}^{2} + 2(1) + 15 = 3 + 2 + 15 = 20$
+    <li> $f(n) = 3n^{2} - 2n + 15$ <br/> 
+    $f(1) = 3{(1)}^{2} - 2(1) + 15 = 3 - 2 + 15 = 16$
+    <li> $f(n) = 3n^{2} + n + 15$ <br/> 
+    $f(1) = 3{(1)}^{2} + 1 + 15 = 3 + 1 + 15 = 19$
+    <li> $f(n) = 3n^{2} - n + 15$ <br/> 
+    $f(1) = 3{(1)}^{2} - 1 + 15 = 3 - 1 + 15 = 17$
 </ul>
+
 Therefore, we know that $f(n) = 3n^{2} - n + 15$ is the correct answer

src/content/questions/comp2804/2015-winter-midterm/11/solution.md

@@ -1,8 +1,11 @@
 We can write down some valid possibilities first.
-$b, B_{n-1}$ possibilities left
-$c, B_{n-1}$ possibilities left
-$a,b, B_{n-2}$ possibilities left
-$a,c, B_{n-2}$ possibilities left
-$a,a,b, B_{n-3}$ possibilities left
-$a,a,c, B_{n-3}$ possibilities left
+
+<ul>
+    <li> $b, B_{n-1}$ possibilities left
+    <li> $c, B_{n-1}$ possibilities left
+    <li> $a,b, B_{n-2}$ possibilities left
+    <li> $a,c, B_{n-2}$ possibilities left
+    <li> $a,a,b, B_{n-3}$ possibilities left
+    <li> $a,a,c, B_{n-3}$ possibilities left
+</ul>
 $S_n = 2 \cdot B_{n-1} + 2 \cdot B_{n-2} + 2 \cdot B_{n-3}$

src/content/questions/comp2804/2015-winter-midterm/12/solution.md

@@ -1,5 +1,9 @@
 Let's suppose we have a string of 1s and 2s.
+
 We can write down some possibilities.
+
 $2, S_{n-2}$ possibilities left
+
 $1,2, S_{n-3}$ possibilities left
+
 $S_n = S_{n-2} + S_{n-3}$

src/content/questions/comp2804/2015-winter-midterm/14/solution.md

@@ -1,15 +1,18 @@
 We can calculate the first couple of values of $S(n)$ to see if we can find a pattern.
+
 $S(1) = 1$
+
 $S(2) = 3$
 
 <ul>
-<li> $S(n)=1+logn$ <br/> 
-$S(2) = 1+log{2} = 1+1 = 2$
-<li> $S(n)=2+nlogn$ <br/> 
-$S(2) = 2+2 log{2} = 2+2 = 4$
-<li> $S(n)=2n+1$ <br/> 
-$S(2) = 2 \cdot 2 + 1 = 5$
-<li> $S(n)=2n-1$ <br/> 
-$S(2) = 2 \cdot 2 - 1 = 3$
+    <li> $S(n)=1+logn$ <br/> 
+    $S(2) = 1+log{2} = 1+1 = 2$
+    <li> $S(n)=2+nlogn$ <br/> 
+    $S(2) = 2+2 log{2} = 2+2 = 4$
+    <li> $S(n)=2n+1$ <br/> 
+    $S(2) = 2 \cdot 2 + 1 = 5$
+    <li> $S(n)=2n-1$ <br/> 
+    $S(2) = 2 \cdot 2 - 1 = 3$
 </ul>
+
 Therefore, we know that $S(n)=2n-1$ is the correct answer.

src/content/questions/comp2804/2015-winter-midterm/15/solution.md

@@ -1,5 +1,9 @@
 $ Pr\text{(number of heads is even)}=1/2 $
+
 $ Pr\text{(number of heads is odd)}=1/2 $
+
 $ Pr\text{(number of tails is even)}=1/2 $
+
 $ Pr\text{(number of tails is odd)}=1/2 $
+
 1/2 is the answer

src/content/questions/comp2804/2015-winter-midterm/17/solution.md

@@ -1,7 +1,13 @@
 We can grab 2 pilsner beers, 1 India pale ale beer, and 1 German wheatbeer beer.
+
 $\binom{7}{2} \cdot 5 \cdot 4$
+
 We can also grab 1 pilsner beer, 2 India pale ale beers, and 1 German wheatbeer beer.
+
 $7 \cdot \binom{5}{2} \cdot 4$
+
 We can also grab 1 pilsner beer, 1 India pale ale beer, and 2 German wheatbeer beers.
+
 $7 \cdot 5 \cdot \binom{4}{2}$
+
 Thus, the total number of ways is $\binom{7}{2} \cdot 5 \cdot 4 + 7 \cdot \binom{5}{2} \cdot 4 + 7 \cdot 5 \cdot \binom{4}{2}$

src/content/questions/comp2804/2015-winter-midterm/2/solution.md

@@ -1,2 +1,3 @@
 For each element in A, there are output 15 choices in B.
+
 Thus, there are $15^7$ functions.

src/content/questions/comp2804/2015-winter-midterm/3/solution.md

@@ -1,2 +1,3 @@
 There are 15 choices for the first element in A, 14 choices for the second element in A, and so on.
+
 Thus, there are $15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9$ one-to-one functions.

src/content/questions/comp2804/2015-winter-midterm/4/solution.md

@@ -1,6 +1,11 @@
 The first letter can be anything. So there are 3 choices.
+
 The second letter cannot be the same as the first letter. So there are 2 choices.
+
 For the third letter, there are 2 choices. Same as the second letter.
+
 We can continue this pattern until the end of the string.
+
 Thus, there are $3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ valid strings.
+
 Rewriting this, we get $V_n = 3 \cdot 2^{n-1}$.

src/content/questions/comp2804/2015-winter-midterm/5/solution.md

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 Let A be the set of strings that start with $abc$.
+
 |A| = first 3 letters are set in stone while other 96 letters can be anything.
+
 Thus, there are $3 \cdot 3 \cdot 3 \cdot 3^{96}$ strings in A.
+
 Let B be the set of strings that end with $bbb$.
+
 |B| = last 3 letters are set in stone while other 96 letters can be anything.
+
 Thus, there are $3^{96} \cdot 3 \cdot 3 \cdot 3$ strings in B.
+
 Let $A \cap B$ be the set of strings that start with $abc$ and end with $bbb$.
+
 $|A \cap B|$ = first 3 letters and last 3 letters are set in stone while other 93 etters can be anything.
+
 $|A \cup B| = |A| + |B| - |A \cap B|$
+
 $|A \cup B| = 3^{96} + 3^{96} - 3^{93}$
+
 $|A \cup B| = 2 \cdot 3^{96} - 3^{93}$

src/content/questions/comp2804/2015-winter-midterm/6/solution.md

@@ -1,6 +1,11 @@
 Let's write it out and see what it represents if n is a small value like 5.
+
 $\sum_{k=2}^{4} (k-1)(5-k)$
+
 $(2-1)(5-2) + (3-1)(5-3) + (4-1)(5-4)$
+
 $1 \cdot 3 + 2 \cdot 2 + 3 \cdot 1$
+
 We can rewrite in \binomial form as
+
 $\binom{n}{3}$

src/content/questions/comp2804/2015-winter-midterm/7/solution.md

@@ -1,5 +1,9 @@
 To guarantee that at least two people share the same birthday, the first 365 people each have a different birthday.
+
 The last person must have a birthday that is the same as one of the previous 365 people.
+
 We can apply this to 3 people.
+
 Thus, the minimum number of people needed is double 365 + 1.
+
 $2 \cdot 365 + 1 = 731$

src/content/questions/comp2804/2015-winter-midterm/8/solution.md

@@ -1,3 +1,5 @@
 $=\sum^{99}_{k=0} \binom{99}{k}17^{99-k}{(2x)}^k$
+
 $=\binom{99}{17}17^{82}{(2x)}^{17}$
+
 $=\binom{99}{17}17^{82}2^{17}x^{17}$

src/content/questions/comp2804/2015-winter-midterm/9/solution.md

@@ -1,3 +1,5 @@
 We can use the dividers method.
+
 We have 99 buckets and 2 extra spots for the dividers to go.
+
 Thus, there are $\binom{99+2}{2}$ solutions.