Task 1 (scheduling problem). In how many ways can a daily school schedule be made, consisting of 6 subjects out of 10?
Solution. If we take, for example, BAGRIF (biology, algebra, geometry, Russian, history, physical education) as one of the possible schedules, then any other schedule must differ from it either in the order of subjects or in the subjects themselves, i.e., we are talking about permutations of six different subjects out of ten. Their number is equal to $A_{10}^{6}=10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5=151200$.

Answer. 151200.

Problem 2 (the number problem). How many different five-digit numbers can be formed from the digits $2,3,5,7$ and 9?

Solution. One of such numbers is 23579. Each new number is obtained from this by permuting its digits. Their number is equal to $P_{5}=5!=$ 120.

Answer. 120.

Problem 3 (the bouquet problem). In how many ways can a bouquet consisting of 5 flowers be composed, given 10 flowers?

Solution. Two bouquets should be considered different if they consist of different flowers, the order does not matter. This means that we are talking about combinations of five flowers out of ten. To determine the number of such bouquets, we reason as follows. Imagine that we have composed all possible bouquets of five flowers, and let the number of these bouquets be $N$. Take one of them and perform all possible permutations in it, the number of which is $P_{5}$. If we perform similar permutations in all bouquets, we get the number of arrangements of 10 out of 5. Thus, $A_{10}^{5}=N P_{5}$. From this, $N=A_{10}^{5}: P_{5}=252$. This number is $C_{10}^{5}$ (see above).

Answer. 252.

The problems are named to associate other problems with these.