A (combinatorics?) problem about shoes "30 shoes are arbitrary ordered in a row, 15 left and 15 right shoes. In this row there will always be 10 succeeding shoes such that 5 of theme are left shoes (and 5 of theme are right shoes. Prove this mathematically"
I suppose this is some kind of combinatorial problem, but I don't manage to prove it.
 A: Let the shoes be numbered from 1 to 30 and let $f(n)$ be the number of left shoes in the set of the $n$-th to the $(n+9)$-th shoe for $1\leq n \leq 21$. $f$ has a minimum that is at most $5$ and a maximum that is at least $5$. Considering the fact that $|f(n+1)-f(n)| \leq 1$ for all $n$ will lead to a proof.
EDIT: Argument with minimum and maximum of $f$.
A: Consider the line of shoes looped to a complete circle, this gives 30 10-shoe series, and since all shoes appear in exactly 10 of those series, the average number of left shoes in these series must be 5.
The neighbouring series of each series only differ by a single shoe, therefore the number of left shoes can only differ by a maximum of 1 between neighbouring series. Therefore if one series has more than 5 left shoes, and another series has less than 5 left shoes, then at least one series in between must have exactly 5 left shoes.
If we are to construct an order that disproves the theorem it must consist of 21 series in a row that either all contain less or all contain more than 5 left shoes, the remaining 9 series must contain so few of many left shoes that the proven average number of left shoes per series is fulfilled.
In the best case scenario 21 of the series contain 4 left shoes each, two of the remaining series are neighbors to one of the 4 left shoe series, so they can contain no more than 5 left shoes, two others of the remaining series are neighbours to these series, so they can contain no more than 6 left shoes etc. When you sum it up there simply cannot be so many left shoes in the 9 series that the average is met, having less than 4 left shoes in any of the 21 series is only going to make the disparity bigger. And there is a similar argument for why 6 left shoes in the 21 series won't work either. Thus it is impossible to create an example that disproves the theorem, and therefore it must be true.
For added fun!
Try to prove the statement for 8, 9, 10, 11 and 12 pairs of shoes. It is false for two of them, and this argument does not work for the remaining three.
