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 possibilities 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 possibilities 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 possibilities 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. 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:

1.1.

6 draws

1.2.

3 wins and 3 losses

1.3.

2 wins, 2 draws, and 2 losses

1.4.

1 win, 4 draws, and 1 loss.

Let’s calculate the number of possibilities for each of these subcases.

1.1.

There is only one way to assign 6 draws to 6 games: The number of possibilities is 1.

1.2.

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

1.3.

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

1.4.

There are $\left(\frac{6}{1}\right)$ ways to assign 1 win to $A$ out of the first 6 games. There are $\left(\frac{5}{4}\right)$ ways to assign 4 draws out of the remaining 5 games. The remaining game is a loss for $A$. The number of possibilities is $\left(\frac{6}{1}\right)\times \left(\frac{5}{4}\right)$.

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:

2.1.

3 wins, 1 draw, and 2 losses

2.2.

2 wins, 3 draws, and 1 loss

2.3.

1 win, 5 draws.

Let’s calculate the number of possibilities for each of these subcases.

2.1.

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 1 draw out of the remaining 3 games. The remaining 2 games are losses for $A$. The number of possibilities is $\left(\frac{6}{3}\right)\times \left(\frac{3}{1}\right)$.

2.2.

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. The remaining game is a loss for $A$. The number of possibilities is $\left(\frac{6}{2}\right)\times \left(\frac{4}{3}\right)$.

2.3.

There are $\left(\frac{6}{1}\right)$ 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 $\left(\frac{6}{1}\right)$.

The total number of possibilities then is:

$$1+\left(\frac{6}{3}\right)+\left(\frac{6}{2}\right)\times \left(\frac{4}{2}\right)+\left(\frac{6}{1}\right)\times \left(\frac{5}{4}\right)+\left(\frac{6}{3}\right)\times \left(\frac{3}{1}\right)+\left(\frac{6}{2}\right)\times \left(\frac{4}{3}\right)+\left(\frac{6}{1}\right)$$