King Arthur has 50 knights sitting at the Round Table, and he wants to send 15 of them to investigate the Holy Grail. In how many ways can he do this if he does not want to select adjacent knights?
If we need to select $k$ knights from a row of $n$ knights such that no two selected knights are adjacent, we can proceed as follows. Among the $n-k$ unselected knights, and at the two ends, there are $n-k+1$ positions. The number of ways to place the $k$ "to be selected" knights into these positions is the number of possible selections, which is $\binom{n-k+1}{k}$.

If we now seat the knights in a row around a round table, then the selections from before are not allowed if both of the end knights are selected. The number of such cases is

$$
\binom{(n-4)-(k-2)+2}{k-2}=\binom{n-k-1}{k-2}
$$

Thus, King Arthur can generally choose in $\binom{n-k+1}{k}-\binom{n-k-1}{k-2}$ ways.

Given the data of the problem, this is $\binom{36}{15}-\binom{34}{13}=4639918800$.

Note. In the general case, the question only makes sense if the selection is actually possible, i.e., if $n-k+1 \geq k$, or $k \leq \frac{n+1}{2}$ for knights in a row, and if $k \leq \frac{n}{2}$ for the round table, since here there is one fewer position between the selected knights.