Problem: I would like to know the number of elements in the cartesian power $X^n$ (cartesian product of one set $X$ by itself, $n$ times) with a maximum constraint: how many elements in $X^n$ have less than $k$ same elements $x$ of the set ($\forall x\in X$)?
Simple example: with $X=\{A, B, C\}$ and $n=3$. The question is: how many three-letter words have less than $k$ times the same letter?
Solution for the simple example:
for $k=4$, there is no constraint: $|X|^n$, here 27 possibilities.
for $k=3$, we just can't have three times the same letter, so $AAA, BBB$ and $CCC$ are forbidden, $27 - 3=25$ possibilities.
for $k=2$, we can't have twice the same letter, so we only have $ABC, ACB, BAC, BCA, CAB, CBA$, 6 possibilities.
for $k=1$, we can't build any word, 0 possibilities.
General solution? What is the general formula to compute this number as a function of $k$?
Strategy 1: all cases minus cases with more than $k$
One solution would be to count all possibilities without constraints, $|X|^n$, and then remove all cases with $k, k+1, ..., n$ repetitions of all elements.
For one given element in $X$
with $k$ of this element: we fix $k$ times the element: $n$ free spaces, $k$ elements, $\binom{n}{k}$ possibilities. For each possibility, we put the others: $n-k$ free spaces, $|X|-1$ elements, $(|X|-1)^{n-k}$ possibilities. In total: $(|X|-1)^{n-k}\cdot \binom{n}{k}$ possibilities for a given other element with $k$ appearances.
For $k+1$ appearances: $(|X|-1)^{n-k-1}\cdot \binom{n}{k+1}$ possibilities
...
For $n$ appearances: $(|X|-1)^{n-n}\cdot \binom{n}{n}= 1$ possibility, intuitive.
So, for one specific other element: $\sum_{j=k}^{n} (|X|-1)^{n-j}\cdot \binom{n}{j}$ possibilities
- For all elements in $X$:
There are $|X|$ elements, so the number of possibilities with more than $k$ times any other element is $|X|\sum_{j=k}^{n} (|X|-1)^{n-j}\cdot \binom{n}{j}$
Problem: some words are double counted
Strategy 1 for the simple example:
All cases: $|X|^n=3^3=27$ possibilities.
Let's fix one element $A\in X$ and iterate over $k$.
Let's fix $k=3$. How many sequences with 3 $A$'s? $\binom{3}{3}=1$ possibility.
Let's fix $k=2$. How many sequences with 2 $A$'s? $\binom{3}{2}=3$ possibilities to put the $A$'s. Then, there is one extra space, that we can fill in with $B$ or $C$: two possibilities. In total: $3\times 2 = 6$ possibilities.
Let's fix $k=1$. How many sequences with 1 $A$? $\binom{3}{1} = 3$ possibilities. Then, there are two extra spaces to fill in with $B$ and $C$. If we accept to have twice $B$, the generated word will be $ABB$, and so it's double counted with "Fix one element $B\in X$, $k=2$".
Strategy 2: (by Andrei Rykhalski)
Construct $X_k$ $\forall k=1, ..., n$ containing $k$ times each element $x\in X$.
Build words of length $n$ from alphabet of size $nk$.
Then find all possible distinct combinations of its elements of length n (exclude duplicates).
In the simple example: $X_3=\{A,A,A,B,B,B,C,C,C\}$, $X_2=\{A,A,B,B,C,C\}$, $X_1=\{A,B,C\}$.
