Mongolian Mathematical Olympiad Round 3

Answer: 126.
Let $1, 2, 3, \ldots, 21$ be points on a circle in clockwise order and let us use the following notation.
$$\begin{aligned} a_1 &= 1, & a_2 &= 8, & a_3 &= 15 \\ b_1 &= 4, & b_2 &= 11, & b_3 &= 18 \\ c_1 &= 7, & c_2 &= 14, & c_3 &= 21 \\ d_1 &= 10, & d_2 &= 17, & d_3 &= 3 \\ e_1 &= 13, & e_2 &= 20, & e_3 &= 6 \\ f_1 &= 16, & f_2 &= 2, & f_3 &= 9 \\ g_1 &= 19, & g_2 &= 5, & g_3 &= 12 \end{aligned}$$

Then we can take only one $a_i$. This means the seven points must be $a_i, b_j, c_k, d_l, e_m, f_n, g_t$. In order to not have 3 unit arc distance, we must have
$i \neq j, j \neq k, k \neq l, l \neq m, m \neq n, n \neq t, t \neq i.$
The total occurrence number of $a_i, b_j, c_k, d_l, e_m, f_n, g_t$ is $3^7$.
Now let us use inclusion-exclusion principle, to count an occurrence numbers at least one index is equal. Let $A_1$ be the set of all occurrences with $i = j$. Let $A_2$ be the set of all occurrences with $j = k$, et cetera, and let $A_7$ be the set of all occurrences with $t = i$. Then we have
$$|A_1 \cup A_2 \cup \cdots \cup A_7| = \sum |A_i| - \sum |A_i \cap A_j| + \cdots - |A_1 \cap A_2 \cap \cdots \cap A_7|.$$
Here we have
$$\begin{align*} \sum |A_i| &= C_7^1 \cdot 3^6 \\ \sum |A_i \cap A_j| &= C_7^2 \cdot 3^5 \\ \sum |A_i \cap A_j \cap A_k| &= C_7^3 \cdot 3^4 \\ \sum |A_i \cap A_j \cap A_k \cap A_l| &= C_7^4 \cdot 3^3 \\ \sum |A_i \cap A_j \cap A_k \cap A_l \cap A_m| &= C_7^5 \cdot 3^2 \\ \sum |A_i \cap A_j \cap A_k \cap A_l \cap A_m \cap A_n| &= C_7^6 \cdot 3 \\ |A_1 \cap A_2 \cap \cdots \cap A_7| &= 3. \end{align*}$$
Hence, the number of configurations satisfying the condition is
$$3^7 - C_7^1 \cdot 3^6 + C_7^2 \cdot 3^5 - C_7^3 \cdot 3^4 + C_7^4 \cdot 3^3 - C_7^5 \cdot 3^2 + C_7^6 \cdot 3 - 3 = (3-1)^7 - 2 = 126.$$