Problem
Given a lottery system where $N$ balls are drawn at random (and not put back in the pool) and you can submit tickets containing between $x_{min}$ and $x_{max}$ predictions, is there a minimum number of tickets $T$ that you can fill out to hit every possible combination of $n$ balls, or every combination of balls up to $n\leq N$?
For example in my country, there's a lottery that uses $45$ consecutively numbered balls of which $6$ are picked at random and then one more as the 'bonus' number. You can submit $6$ to $10$ predictions per ticket at varying costs. Depending on the number of correct predictions and whether you got the bonus ball right or not, you can win different amounts of money.
Example
Let's say $N=5$, $x_{min} = 2$ and $x_{max}=3$. Two balls are drawn at random, as well as a bonus number.
Single number combinations
There are ${5 \choose 1} = 5$ combinations which you can make with $T=2$ tickets as $(1,2,3) \cup (4,5)$. You could also buy $T=3$ tickets, two containing two different pairs of numbers and one containing the remaining number and a duplicate number.
Two-number combinations
There are ${5 \choose 2} = 10$ combinations which using a greedy algorithm can be made with $T=4$ tickets as $(3,2,4) \cup (3,1,5) \cup (4,2,5) \cup (4,1,2)$ in no particular order. You could also buy $T=10$ tickets and fill in a pair each.
$n$-number combinations generalized?
For $n=1$ and $n=2$, above, the formula $T(N,x_{max},n)=\lceil {N \choose n} / {x_{max} \choose n} \rceil$ yields the number of tickets I'm thinking of. If this generally extends, then that would mean I would need $T(45,10,6)= \lceil {45 \choose 6} / {10 \choose 6} \rceil = 38 \, 786$ tickets. But does it?
Two-number and bonus number combinations
Since there are $10$ possible two-number combinations and $3$ remaining balls that can be the bonus number, that should be $30$ pair-bonus combinations right? I don't quite see how I can combine this idea with the general formula for $T$ above.
My goal
Given all possible ticket combinations for each required number of correct guesses, I can use code to determine which is the cheapest and also how much money I would make (or probably lose). I'm also not sure if having $5$ correct numbers automatically also means that you got $4$, $3$, and $2$ right as well.
How can I extend this idea to higher values of $N$, $x_{min}$, $x_{max}$ and $n$?