1. Oleg has 550 rubles, and he wants to give his mother tulips for March 8, and there must be an odd number of them, and no color shade should be repeated. In the store where Oleg came, one tulip costs 49 rubles, and there are eleven shades of flowers available. How many ways are there for Oleg to give his mother flowers? (The answer in the problem should be a compact expression, without summation signs, ellipses, etc.)
Solution. From the condition, it is obvious that the maximum number of flowers in a bouquet is 11.

1st method

Using the property of binomial coefficients

$$
\mathrm{C}_{n}^{1}+\mathrm{C}_{n}^{3}+\mathrm{C}_{n}^{5}+\cdots=2^{n-1}
$$

and also considering their combinatorial meaning, we get that the number of ways to form a bouquet from an odd number of flowers of no more than 11 shades (with the condition that no shade should be repeated) is:

$$
\mathrm{C}_{11}^{1}+\mathrm{C}_{11}^{3}+\mathrm{C}_{11}^{5}+\cdots+\mathrm{C}_{11}^{11}=2^{10}=1024
$$

2nd method

Consider 10 flowers of 10 different shades. A bouquet can be assembled from these flowers without considering the parity in $2^{10}$ ways. If the bouquet has an odd number of flowers, we keep it; if it has an even number, we add the unused eleventh flower. Thus, the total number of ways to assemble a bouquet is $2^{10}$.

Answer: 1024.