3. Maximum 15 points. Masha was given a chest with multicolored beads (each bead has a unique color, there are a total of $\mathrm{n}$ beads in the chest). Masha chose seven beads for her dress and decided to try all possible combinations of them on the dress (thus, Masha selects from a set of options to sew one, two, three, four, five, six, or seven beads, and the order of the beads does not matter to her). Then she counted how many options she got and was very surprised that the number turned out to be odd.
1) What number did Masha get?
2) Is it true that if Masha had chosen from an even number of beads, she could have gotten an even number of options?
3) Is it true that if the order of the beads sewn on the dress mattered to Masha, she could have gotten both even and odd numbers?
# Solution:

1) Consider one bead. Before Masha sews it onto the dress, there are two options: to take the bead or not. If we now choose two beads, the number of options becomes four, which can be obtained by multiplying the first option by two. By increasing the number of beads, we conclude that the total number of all possible options for p beads is $2^{\text {n }}$. According to the problem, Masha chose 7 beads, but the option of not sewing anything is not considered, so the number of options is $2^{7}-1=127$.
2) The number of options is odd (see point 1).
3) This is correct; it is sufficient to consider the case of one and two beads. In the first case, the number of options is 1, and in the second case, it is 4.

| Criterion | Score |
| :--- | :---: |
| The correct answer is obtained in points 1), 2), and 3) through a logically justified algorithm | 15 |
| The correct answer is obtained in only one of the points through a logically justified algorithm | 2 |
| The solution does not meet any of the presented criteria | 0 |