Placing balls
Task number: 3449
We wish to place \(k\) balls into \(n\) pigeonholes. Fill in the following table with the numbers of possible choices:
Balls have |
Each pigeonhole containse | ||
---|---|---|---|
at most one | any number | at least one | |
distinct colors | |||
the same color |
Solution
Choosing balls with distinct colors corresponds to a mapping \(f\), where \(f(i)\) indicates the pigeonhole into which to place the \(i\)-th ball. If each pigeonhole may contain at most one ball, this must be an injective mapping.
We count surjective mappings using the inclusion-exclusion principle, i.e. by (more easily) counting the mappings which are not surjective. If \(A_j\) indicates the set of mappings \(f\) where the element \(j\) is not in the image of \(f\), we obtain \(\left|\bigcap\limits_{j\in I} A_j\right|=(n-|I|)!\). The resulting equation is an application of the inclusion-exclusion principle to the intersections with the index set \(I\) of size \(i\).
If we are choosing balls of a single color and at most one ball will fit into each pigeonhole, this corresponds to a choice of a \(k\)-element subset of pigeonholes, where the pigeonholes in the chosen set will be the ones with balls.
If more balls can fit into a pigeonohole, this corresponds to the sum \(x_1+…+x_n=k\), where \(x_i\) denotes the number of balls in the \(i\)-th pigeonhole.
If each pigeonhole must contain at least one ball, we first separate \(n\) balls, place one in each pigeonhole and distribute the remaining balls as in the preceding case.
Answer
Balls
haveEach pigeonhole containse at most one any number at least one distinct colors \(\frac{n!}{(n-k)!}\) \(n^k\) \(\sum\limits_{i=0}^{n-1} (-1)^i \binom{n}{i} (n-i)^k \) the same color \(\binom{n}{k}\) \(\binom{n+k-1}{k}\) \(\binom{k-1}{n-1}\)