In a certain lottery, there are 60 distinct numbers and 6 of them are drawn without replacement. Each ticket has 6 distinct numbers among the 60 possible ones. The top prize, known as the "goal-in-the-corner," is awarded to the player who has the ticket with the same 6 numbers that were drawn. In this lottery, there is also the "ball-on-the-post" prize. In a ball-on-the-post ticket, the smallest number does not have a difference, in absolute value, greater than 1 from the smallest number drawn, the second smallest number does not have a difference, in absolute value, greater than 1 from the second smallest number drawn, and so on until the sixth smallest number. For example, suppose the goal-in-the-corner ticket is $\{4,7,25,48,51,60\}$. Then the tickets $\{3,6,25,49,50,59\}$ and $\{5,6,25,47,50,60\}$ are ball-on-the-post tickets, but the ticket $\{3,4,6,24,47,50\}$ is not a ball-on-the-post ticket. It is worth noting that a goal-in-the-corner ticket is not a ball-on-the-post ticket. For the following items, consider each ticket as the selection of a sequence of 6 numbers written in ascending order.

(a) Give an example of a set of 6 numbers forming a goal-in-the-corner ticket that has the smallest number of ball-on-the-post tickets associated with it. How many ball-on-the-post tickets would there be for these 6 numbers?

(b) Give an example of a set of 6 numbers forming a goal-in-the-corner ticket that results in the largest number of ball-on-the-post tickets. In this case, how many ball-on-the-post tickets would there be?

(c) Consider the numbers drawn $\{2,3,8,11,14,17\}$, how many ball-on-the-post tickets are associated with it?

(d) Suppose the set of numbers drawn is $\{8,10,12,14,16,18\}$. In this case, how many ball-on-the-post tickets are possible associated with it?
Solution

(a) Take the 6 smallest numbers $\{1,2,3,4,5,6\}$. In this case, we have only 6 bolana-trave tickets, which are formed by choosing 6 numbers from the set $\{1,2,3,4,5,6,7\}$. In principle, there would be 7 tickets, but since a gol-no-ângulo ticket is not a bolana-trave, the number of bolana-trave tickets associated with it is 6. We will prove that this is the smallest possible value by showing that all other sets have at least 6 bolana-trave tickets.

Taking the set $\{a, b, c, d, e, f\}$ in ascending order from $a$ to $f$. If $f>1$, we have the tickets:

$$
\begin{gathered}
\{a-1, b, c, d, e, f\} \\
\{a-1, b-1, c, d, e, f\} \\
\{a-1, b-1, c-1, d, e, f\} \\
\{a-1, b-1, c-1, d-1, e, f\} \\
\{a-1, b-1, c-1, d-1, e-1, f\} \\
\{a-1, b-1, c-1, d-1, e-1, f-1\}
\end{gathered}
$$

If $a=1$ and $f=60$, then there would be two consecutive numbers that are not consecutive, for example, $d-c \geq 2$, in this case we would have the tickets:

$$
\begin{gathered}
\{a, b, c+1, d, e, f\} \\
\{a, b+1, c+1, d, e, f\} \\
\{a+1, b+1, c+1, d, e, f\} \\
\{a, b, c, d-1, e, f\} \\
\{a, b, c, d-1, e-1, f\} \\
\{a, b, c, d-1, e-1, f-1\}
\end{gathered}
$$

(b) See that each number $x$ in the gol-no-ângulo ticket generates 3 possibilities for its position in the bolana-trave, namely $x-1, x$ and $x+1$. If consecutive numbers are close, then we will have to eliminate repetitions, for example, if $a=b-1$ or $a+1=b-1$. Therefore, to reach the maximum, we can leave consecutive numbers with a difference of at least 3. An example is the gol-no-ângulo ticket $\{3,6,9,12,15,18\}$. A bolana-trave ticket has three possibilities for the smallest number, three for the second smallest, and so on. Remembering not to count the gol-no-ângulo ticket, we will have a total of $3^{6}-1=728$ bolana-trave tickets associated with it.
(c) Let's separate into cases. If the first number is 1, then for each of the five numbers we have three possibilities, resulting in $3^{5}$ tickets. If the first number is 2, then the second number must be 3 or 4. The others have three possibilities each, so we will have $2 \cdot 3^{4}$ bolana-trave tickets associated. If the first number is 3, the second is necessarily 4. For each of the others, there are three possibilities. So we will have $3^{4}$ possibilities. Remembering to remove the gol-no-ângulo ticket, we are left with

$$
3^{5}+2 \cdot 3^{4}+3^{4}-1=485
$$

bolana-trave tickets associated with the gol-no-ângulo $\{2,3,8,11,14,17\}$.

(d) Let's start by analyzing the sets $\{8,10,12\}$ and $\{14,16,18\}$ separately. For the first group of numbers, if the largest number in the bolana-trave is 13, there are three possibilities for 08 and three for the 10, with the exception of both choices being equal to 9. So there are eight possibilities in this case. If the largest number is 12, we also have 8 possibilities, and if the largest is 11, we have only two possibilities for 10 and three for 08, with the exception of both choices being equal to 9. So we have 5 possibilities. Thus, for the first group, we have $8+8+5=21$ possibilities. By symmetry, we can observe the possibilities of the smallest number of the second group and conclude that for the second group we also have $8+8+5=21$ possibilities. See that we have to subtract the possibilities where 13 was chosen for the 12 and 13 also for the 14, that is, $8 \cdot 8=64$ possibilities. Again, we have to remember to exclude the gol-no-ângulo ticket. So we will have

$$
21 \cdot 21-8 \cdot 8-1=441-64-1=376
$$

bolana-trave tickets associated with $\{8,10,12,14,16,18\}$.