A1. How many numbers are the sum of seven different numbers from the numbers $0,1,2,3$, $4,5,6,7,8$ and 9?
A1. The smallest number you can make as the sum of seven numbers is:

$0+1+2+3+4+5+6=21$. By replacing the last number each time, you get:

$0+1+2+3+4+5+7=22$,

$0+1+2+3+4+5+8=23$,

$0+1+2+3+4+5+9=24$. Then replace the second-to-last number:

$0+1+2+3+4+6+9=25 \ldots$ and so on. The last and largest number becomes:

$3+4+5+6+7+8+9=42$, with all the numbers in between, thus $21 \mathrm{t} / \mathrm{m} 42$.

In total, that's 22 numbers.

![](https://cdn.mathpix.com/cropped/2024_04_17_774e242ba19fa9593e58g-1.jpg?height=149&width=1596&top_left_y=708&top_left_x=230)

With two colors, you have 3 possibilities regarding the colors. For each of those

![](https://cdn.mathpix.com/cropped/2024_04_17_774e242ba19fa9593e58g-1.jpg?height=91&width=1468&top_left_y=908&top_left_x=228)

In total, that's 12 possibilities with two colors.

With three colors, you have three possibilities regarding the distribution of the colors over the beads, because you must have two beads of the same color. With the combination $r r w b$, you have the following two bracelets: $\begin{array}{ll}r & r \\ b & w\end{array}$ (be careful: $\begin{array}{ll}r & r \\ w & b\end{array}$ is the same!) and $\begin{array}{ll}r & w \\ b & r\end{array}\left(\begin{array}{ll}r & b \\ w & r\end{array}\right)$. So with three colors, in total 6. All possibilities together: $3+12+6=21$.