## Task 2 - 250832

Brigade Schulz plays "Tele-Lotto (5 out of 35)" according to a so-called "fully mathematical system with $n$ numbers". This means that if $n$ is a natural number with $5 < n \leq 35$, the system is as follows:

Exactly $n$ numbers from 1, 2, ..., 35 are selected, and then all different subsets of five elements are determined in the set of these $n$ numbers. Each of these subsets is submitted as a tip.

a) The brigade plays according to a "fully mathematical system with 6 numbers". In the subsequent draw in Tele-Lotto, it turns out that exactly four of the five winning numbers are among the six numbers used by the brigade in their system.

Furthermore, the following prize amounts are announced after this draw: For each tip with exactly three correctly predicted numbers, there is $20 \mathrm{M}$, for each tip with exactly four correctly predicted numbers, there is $400 \mathrm{M}$, and for each tip with five correctly predicted numbers, there is $3000 \mathrm{M}$.

Determine the winnings of the brigade in this case!

b) The brigade plays according to a "fully mathematical system with 7 numbers". In a draw, it is observed that exactly four of the five winning numbers are among the seven numbers used by the brigade. The prize amounts are the same as in a).

Determine the winnings of the brigade in this case as well!

Hint: The costs, i.e., the game stake, should not be considered in both tasks.
a) Among the six numbers used by the brigade, there are exactly six different subsets of five elements each; these subsets arise by omitting one of the six numbers each time.

Exactly two of the six numbers are not winning numbers. Therefore, there are exactly the following two types (1), (2) among the tips submitted by the brigade:

(1) Tips with exactly one number that is not a winning number,

(2) Tips with the two numbers that are not winning numbers.

There are exactly two tips of type (1), namely those in which the four winning numbers and one of the other two numbers appear.

Consequently, the four other tips of the system are exactly of type (2). Thus, the winnings amount to $2 \cdot 400 M + 4 \cdot 20 M = 880 M$.

b) Among the seven numbers used by the brigade, there are exactly 21 different subsets of five elements each.

These subsets arise by omitting exactly two of the seven numbers each time, and this can be done in the following ways:

- either the first of the seven numbers and one of the remaining six (six possibilities),
- or the second and one of the five numbers different from the first and second (five possibilities),
- or the third and one of the four numbers different from the first to third (four possibilities),
- or the fourth and one of the three numbers different from the first to fourth (three possibilities),
- or the fifth and one of the two numbers different from the first to fifth (two possibilities),
- or the sixth and the seventh number (one possibility);

the number of these subsets is thus $6+5+4+3+2+1=21$.

Exactly three of the seven numbers are not winning numbers. Therefore, there are exactly the following three types (1), (2), (3) among the tips submitted by the brigade:

(1) Tips with exactly one number that is not a winning number,

(2) Tips with exactly two numbers that are not winning numbers,

(3) Tips with the three numbers that are not winning numbers.

There are exactly three tips of type (1), namely those in which the four winning numbers and one of the other three numbers appear. There are exactly six tips of type (3), namely those in which, in addition to the three numbers that are not winning numbers, two of the four winning numbers appear, and these are either the first and one of the other three winning numbers (three possibilities) or the second and one of the two different from the first and second winning numbers (two possibilities) or the third and fourth winning numbers (one possibility).

Consequently, the other $21-3-6=12$ tips of the system are of type (2).

Thus, the winnings amount to $3 \cdot 400 M + 12 \cdot 20 M = 1440 M$.