fix chapter 1 problem 7 solution

Author: michrzan

src/chapters/1/sections/counting/problems/7.tex

@@ -1,89 +1,91 @@
 \begin{enumerate}[label=(\alph*)]
-\item There are ${7 \choose 3}$ ways to assign three wins to player $A$. For a 
-specific combination of three games won by $A$, there are ${4 \choose 2}$ ways 
-to assign two draws to $A$. There is only one way to assign two losses to $A$ 
-from the remaining two games, namely, $A$ losses both games.
-
-  $$ {7 \choose 3} \times {4 \choose 2} \times {2 \choose 2} $$
-
-\item   If $A$ were to draw every game, there would need to be at least $8$ 
-games for $A$ to obtain $4$ points, so $A$ has to win at least $1$ game. 
-Similarly, if $A$ wins more than $4$ games, they will have more than $4$ points.
-  
-  Case 1: $A$ wins $1$ game and draws $6$.
-  
-  This case amounts to selecting $1$ out of $7$ for $A$ to win and assigning a 
-  draw for the other $6$ games. Hence, there are $7$ possibilities.
+  \item There are ${7 \choose 3}$ ways to assign three wins to player $A$. For a 
+  specific combination of three games won by $A$, there are ${4 \choose 2}$ ways 
+  to assign two draws to $A$. There is only one way to assign two losses to $A$ 
+  from the remaining two games, namely, $A$ losses both games.
+  
+    $$ {7 \choose 3} \times {4 \choose 2} \times {2 \choose 2} $$
+  
+  \item   If $A$ were to draw every game, there would need to be at least $8$ 
+  games for $A$ to obtain $4$ points, so $A$ has to win at least $1$ game. 
+  Similarly, if $A$ wins more than $4$ games, they will have more than $4$ points.\\
+    \\
+    Case 1: $A$ wins $1$ game and draws $6$.
 
-  Case 2: $A$ wins $2$ games and draws $4$.
-  
-  There are ${7 \choose 2}$ ways to assign $2$ wins to $A$. For each of them, 
-  there are ${5 \choose 4}$ ways to assign four draws to $A$ out of the 
-  remaining $5$ games. Player $B$ wins the remaining game. The total number 
-  of possibilites for this case is ${7 \choose 2} \times {5 \choose 4}$.
-  
-  Case 3: $A$ wins $3$ games and draws $2$.
-  
-  There are ${7 \choose 3}$ ways to assign $3$ wins to $A$. For each of them, 
-  there are ${4 \choose 2}$ ways to assign two draws to $A$ out of the remaining 
-  $4$ games. $B$ wins the remaining $2$ games. The total number of possibilites 
-  for this case is ${7 \choose 3} \times {4 \choose 2}$.
-  
-  Case 4: $A$ wins $4$ games and loses $3$.
-  
-  There are ${7 \choose 4}$ ways to assign $4$ wins to $A$. $B$ wins the 
-  remaining $3$ games.
-  The total number of possibilites for this case is ${7 \choose 4}$.
-  
-  Summing up the number of possibilities in each of the cases we get
-  
-  $$ {7 \choose 1} + {7 \choose 2} \times {5 \choose 4} + {7 \choose 3} \times 
-  {4 \choose 2} + {7 \choose 4} $$
-
-\item If $B$ were to win the last game, that would mean that $A$ had already 
-obtained $4$ points prior to the last game, so the last game would not be 
-played at all. Hence, $B$ could not have won the last game.
-  
-  Case 1: $A$ wins $3$ out of the first $6$ games and wins the last game.
-  
-  There are ${6 \choose 3}$ ways to assign $3$ wins to $A$ out of the first $6$ 
-  games. The other $3$ games end in a draw. The number of possibilities then 
-  is ${6 \choose 3}$.
-  
-  Case 2: $A$ wins $2$ and draws $2$ out of the first $6$ games and wins the 
-  last game.
-  
-  There are ${6 \choose 2}$ ways to assign $2$ wins to $A$ out of the first 
-  $6$ games. From the $4$ remaining games, there are ${4 \choose 2}$ ways to 
-  assign $2$ draws. The remaining $2$ games are won by $B$. The number of 
-  possibilities is ${6 \choose 2} \times {4 \choose 2}$.
+    This case amounts to selecting $1$ out of $7$ for $A$ to win and assigning a 
+    draw for the other $6$ games. Hence, there are $7$ possibilities.
+      
+    Case 2: $A$ wins $2$ games and draws $4$.
+    
+    There are ${7 \choose 2}$ ways to assign $2$ wins to $A$. For each of them, 
+    there are ${5 \choose 4}$ ways to assign four draws to $A$ out of the 
+    remaining $5$ games. Player $B$ wins the remaining game. The total number 
+    of possibilities for this case is ${7 \choose 2} \times {5 \choose 4}$.
+    
+    Case 3: $A$ wins $3$ games and draws $2$.
+    
+    There are ${7 \choose 3}$ ways to assign $3$ wins to $A$. For each of them, 
+    there are ${4 \choose 2}$ ways to assign two draws to $A$ out of the remaining 
+    $4$ games. $B$ wins the remaining $2$ games. The total number of possibilities 
+    for this case is ${7 \choose 3} \times {4 \choose 2}$.
+    
+    Case 4: $A$ wins $4$ games and loses $3$.
+    
+    There are ${7 \choose 4}$ ways to assign $4$ wins to $A$. $B$ wins the 
+    remaining $3$ games.
+    The total number of possibilities for this case is ${7 \choose 4}$.
+    
+    Summing up the number of possibilities in each of the cases we get
+    
+    $$ {7 \choose 1} + {7 \choose 2} \times {5 \choose 4} + {7 \choose 3} \times 
+    {4 \choose 2} + {7 \choose 4} $$
+  
+  \item If $B$ were to win the last game, that would mean that $A$ had already 
+  obtained $4$ points prior to the last game, so the last game would not be 
+  played at all. Hence, $B$ could not have won the last game. 
+  The last game must have ended in either A winning (case 1) or a draw (case 2).\\
+    \\
+    Case 1: $A$ wins the last game. This means $A$ had 3 points
+    after 6 games. 
+    
+    There are four possibilities for $A$ to earn 3 points in 6 games:
 
-  Case 3: The last game ends in a draw.
+    \begin{enumerate}[label=1.\arabic*.]
+      \item 6 draws
+      \item 3 wins and 3 losses
+      \item 2 wins, 2 draws, and 2 losses
+      \item 1 win, 4 draws, and 1 loss.
+    \end{enumerate}
 
-  This case implies that $A$ had $3.5$ and $B$ had $2.5$ points by the end of 
-  game $6$.
+    Let's calculate the number of possibilities for each of these subcases.
 
-  Case 3.1: $A$ wins $3$ and draws $1$ out of the first $6$ games.
+    \begin{enumerate}[label=1.\arabic*.]
+      \item There is only one way to assign 6 draws to 6 games: The number of possibilities is 1.
+      \item There are ${6 \choose 3}$ ways to assign 3 wins to $A$ out of the first 6 games. The remaining 3 games are losses for $A$. The number of possibilities is ${6 \choose 3}$.
+      \item There are ${6 \choose 2}$ ways to assign 2 wins to $A$ out of the first 6 games. There are ${4 \choose 2}$ ways to assign 2 draws out of the remaining 4 games. The remaining 2 games are losses for $A$. The number of possibilities is ${6 \choose 2} \times {4 \choose 2}$.
+      \item There are ${6 \choose 1}$ ways to assign 1 win to $A$ out of the first 6 games. There are ${5 \choose 4}$ ways to assign 4 draws out of the remaining 5 games. The remaining game is a loss for $A$. The number of possibilities is ${6 \choose 1} \times {5 \choose 4}$.
+    \end{enumerate}
 
-  There are ${6 \choose 3}$ ways to assign $3$ wins to $A$ out of the first 
-  $6$ games. There are ${3 \choose 1}$ ways to assign a draw out of the 
-  remaining $3$ games. $B$ wins the other $2$ games. The number of 
-  possibilities is ${6 \choose 3} \times {3 \choose 1}$.
+    Case 2: The last game ends in a draw. This means $A$ had 3.5 points after 6 games. 
+    
+    There are three possibilities for $A$ to earn 3.5 points in 6 games:
 
-  Case 3.2: $A$ wins $2$ and draws $3$ out of the first $6$ games.
-   
-  There are ${6 \choose 2}$ ways to assign $2$ wins to $A$ out of the 
-  first $6$ games. There are ${4 \choose 3}$ ways to assign $3$ draws out of 
-  the remaining $4$ games. $B$ wins the remaining game. The number of 
-  possibilities is ${6 \choose 2} \times {4 \choose 3}$.
+    \begin{enumerate}[label=2.\arabic*.]
+      \item 3 wins, 1 draw, and 2 losses
+      \item 2 wins, 3 draws, and 1 loss
+      \item 1 win, 5 draws.
+    \end{enumerate}
 
-  Case 3.3: $A$ wins $1$ and draws $5$ of the first $6$ games.
+    Let's calculate the number of possibilities for each of these subcases.
 
-  There are ${6 \choose 1}$ ways to assign a win to $A$ out of the first 
-  $6$ games.
+    \begin{enumerate}[label=2.\arabic*.]
+      \item There are ${6 \choose 3}$ ways to assign 3 wins to $A$ out of the first 6 games. There are ${3 \choose 1}$ ways to assign 1 draw out of the remaining 3 games. The remaining 2 games are losses for $A$. The number of possibilities is ${6 \choose 3} \times {3 \choose 1}$.
+      \item There are ${6 \choose 2}$ ways to assign 2 wins to $A$ out of the first 6 games. There are ${4 \choose 3}$ ways to assign 3 draws out of the remaining 4 games. The remaining game is a loss for $A$. The number of possibilities is ${6 \choose 2} \times {4 \choose 3}$.
+      \item There are ${6 \choose 1}$ ways to assign 1 win to $A$ out of the first 6 games. The remaining 5 games are losses for $A$. The number of possibilities is ${6 \choose 1}$.
+    \end{enumerate}
 
-  The total number of possibilities then is
+    The total number of possibilities then is:
+    
+    $$ 1 + {6 \choose 3} + {6 \choose 2} \times {4 \choose 2} + {6 \choose 1} \times {5 \choose 4} + {6 \choose 3} \times {3 \choose 1} + {6 \choose 2} \times {4 \choose 3} + {6 \choose 1} $$
 
-  $$ {6 \choose 3} + {6 \choose 2} \times {4 \choose 2} + {6 \choose 3} \times 
-  {3 \choose 1} + {6 \choose 2} \times {4 \choose 3} + {6 \choose 1}$$
-\end{enumerate}
+  \end{enumerate}