1.1.7 problem 7

(a)

There are $\left(\frac{7}{3}\right)$ ways to assign three wins to player $A$. For a specific combination of three games won by $A$, there are $\left(\frac{4}{2}\right)$ 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.

$$\left(\frac{7}{3}\right)\times \left(\frac{4}{2}\right)\times \left(\frac{2}{2}\right)$$

(b)

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.

Case 2: $A$ wins $2$ games and draws $4$.

There are $\left(\frac{7}{2}\right)$ ways to assign $2$ wins to $A$. For each of them, there are $\left(\frac{5}{4}\right)$ 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 $\left(\frac{7}{2}\right)\times \left(\frac{5}{4}\right)$.

Case 3: $A$ wins $3$ games and draws $2$.

There are $\left(\frac{7}{3}\right)$ ways to assign $3$ wins to $A$. For each of them, there are $\left(\frac{4}{2}\right)$ 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 $\left(\frac{7}{3}\right)\times \left(\frac{4}{2}\right)$.

Case 4: $A$ wins $4$ games and loses $3$.

There are $\left(\frac{7}{4}\right)$ ways to assign $4$ wins to $A$. $B$ wins the remaining $3$ games. The total number of possibilites for this case is $\left(\frac{7}{4}\right)$.

Summing up the number of possibilities in each of the cases we get

$$\left(\frac{7}{1}\right)+\left(\frac{7}{2}\right)\times \left(\frac{5}{4}\right)+\left(\frac{7}{3}\right)\times \left(\frac{4}{2}\right)+\left(\frac{7}{4}\right)$$

(c)

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 $\left(\frac{6}{3}\right)$ 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 $\left(\frac{6}{3}\right)$.

Case 2: $A$ wins $2$ and draws $2$ out of the first $6$ games and wins the last game.

There are $\left(\frac{6}{2}\right)$ ways to assign $2$ wins to $A$ out of the first $6$ games. From the $4$ remaining games, there are $\left(\frac{4}{2}\right)$ ways to assign $2$ draws. The remaining $2$ games are won by $B$. The number of possibilities is $\left(\frac{6}{2}\right)\times \left(\frac{4}{2}\right)$.

Case 3: The last game ends in a draw.

This case implies that $A$ had $3.5$ and $B$ had $2.5$ points by the end of game $6$.

Case 3.1: $A$ wins $3$ and draws $1$ out of the first $6$ games.

There are $\left(\frac{6}{3}\right)$ ways to assign $3$ wins to $A$ out of the first $6$ games. There are $\left(\frac{3}{1}\right)$ ways to assign a draw out of the remaining $3$ games. $B$ wins the other $2$ games. The number of possibilities is $\left(\frac{6}{3}\right)\times \left(\frac{3}{1}\right)$.

Case 3.2: $A$ wins $2$ and draws $3$ out of the first $6$ games.

There are $\left(\frac{6}{2}\right)$ ways to assign $2$ wins to $A$ out of the first $6$ games. There are $\left(\frac{4}{3}\right)$ ways to assign $3$ draws out of the remaining $4$ games. $B$ wins the remaining game. The number of possibilities is $\left(\frac{6}{2}\right)\times \left(\frac{4}{3}\right)$.

Case 3.3: $A$ wins $1$ and draws $5$ of the first $6$ games.

There are $\left(\frac{6}{1}\right)$ ways to assign a win to $A$ out of the first $6$ games.

The total number of possibilities then is

$$\left(\frac{6}{3}\right)+\left(\frac{6}{2}\right)\times \left(\frac{4}{2}\right)+\left(\frac{6}{3}\right)\times \left(\frac{3}{1}\right)+\left(\frac{6}{2}\right)\times \left(\frac{4}{3}\right)+\left(\frac{6}{1}\right)$$