Generalized pigeonhole principle

Statement

Let $A$ and $B$ be finite sets, let $f : A \to B$ be a function, and let $n$ be a natural number. Assume that $|B| \cdot n < |A|$. Then there is an element $b \in B$ whose fibre $f^{-1}(b) = \{a \in A : f(a) = b\}$ has more than $n$ elements.

Remarks

For $n = 1$ this is the pigeonhole principle. An equivalent form: if $N$ objects go into $k$ boxes, some box has at least $\lceil N / k \rceil$ objects. Example: among $25$ people, at least $3$ share a birth month, because $12 \cdot 2 < 25$.

Explanations

L1 human

If every box held at most $n$ objects, the boxes would hold at most $n$ times the number of boxes. So if there are more objects than that, some box must hold more than $n$.