Partial Derangement formula for permutation with repeated elements

Question

MY question is to get general formula for repeated permutation: For any $n$ numbers,

$n=1,2,3, \ldots$

Derangement formula: $$D_n=!n=(n−1)(!(n−1)+!(n−2))$$ Here the numbers are distinct from one another (no repetition of any number in permutation) https://en.wikipedia.org/wiki/Derangement

Partial derangement: Instead of $n$ derangement we have $k$ derangements, for $n \geq 0$ and $0 \leq k \leq n$, the rencontres number $D_{n, k}$ Partial derangement or rencontre number: https://en.wikipedia.org/wiki/Rencontres_numbers

Is there any general formula for partial derangement of permutation with repeated number (repeated numbers exist in permutation). For example:

$n=1,1,2,2,3,3,4,5$

Any general formula for Derangement of $k$ numbers??

Rewriting your above example: suppose A is blue and B,C are red; we have the permutations: \begin{matrix} ABC\rightarrow ABC \\ ABC\rightarrow ACB\\ ABC\rightarrow BAC \\ ABC\rightarrow BCA \\ ABC\rightarrow CAB \\ ABC\rightarrow CBA \\ \end{matrix} For example, we have $N=3$,$M=2$ ($1<M<N$) Trying to calculate Probability : Example -1: $P(\overline{A \ or \ B} ) $ , Results: $\frac{3}{6}$

Similarly Example -2: $P(\overline{A \ or \ C}) $

$P$: Probability, $\overline{A \ or \ B}$ :not hit A or B and so on. Any generalized form of formula to calculate above probability?? I tried with inclusion and exclusion principle but not sure.

Another bigger scenario: suppose A is blue, B is red, C,D are green; We get final polynomial: $2x^4+10x^2+8x+4$ We have the permutations: \begin{matrix} ABCD\rightarrow ABCD (hit-4) \\ ABCD\rightarrow ABDC (hit-4)\\ ABCD\rightarrow ACBD (hit-2) \\ ABCD\rightarrow ACDB (hit-2) \\ ABCD\rightarrow ADBC (hit-2) \\ ABCD\rightarrow ADCB (hit-2) \\ ABCD\rightarrow BACD (hit-2) \\ ABCD\rightarrow BADC (hit-2) \\ ABCD\rightarrow BCAD (hit-1) \\ ABCD\rightarrow BCDA (hit-1) \\ ABCD\rightarrow BCAD (hit-1) \\ ABCD\rightarrow BCDA (hit-1) \\ ABCD\rightarrow CABD (hit-1) \\ ABCD\rightarrow CADB (hit-1) \\ ABCD\rightarrow CBAD (hit-2) \\ ABCD\rightarrow CBDA (hit-2) \\ ABCD\rightarrow CDAB (hit-0) \\ ABCD\rightarrow CDBA (hit-0) \\ ABCD\rightarrow DABC (hit-1) \\ ABCD\rightarrow DACB (hit-1) \\ ABCD\rightarrow DBAC (hit-2) \\ ABCD\rightarrow DBCA (hit-2) \\ ABCD\rightarrow DCAB (hit-0) \\ ABCD\rightarrow DCBA (hit-0) \\ \end{matrix}

For example, we have $N=4$,$M=3$ (any number less than $N$). Trying to calculate Probability : Example -1: $P(\overline{A \ or \ B \ or \ C}) $ , Results $\frac{something}{24}=?/24$

Similarly Example -2: we have $N=4$,$M=2$ ($1<M<N$). Trying to calculate Probability : $P(\overline{A \ or \ C}) $.

Inclusion exclusion principle: $P(A \ or \ B \ or C) $ =$P(A)+ P(B) + P(C) -P(A \cap B) - P(A \cap C) -P(B \cap C) + P(A\cup B \cup C) $. Just trying to get formula to calculate the probability of any number of $N$ and $M$ which will become complex for large number of $N$ and $M$!!!! Any generalized form of formula to calculate above probability from the rook polynomial theory??

I think I can rewrite the problem according to your statement: Given a set $S$ of $n_1+n_2+⋯+n_k$ distinguishable, colored objects, with $n_i$ of them colored with the ith color, how many permutations are there of $S$ so that either any of $r$ elements ($r<=k$) map to their own color (or does not map their own color)?

Answer 1

I am going to assume your problem is the following:

Given a set $S$ of $n_1 + n_2 + \cdots + n_k$ distinguishable, colored objects, with $n_i$ of them colored with the $i$th color, how many permutations are there of $S$ so that exactly $k$ elements map to their own color?

You can solve this problem with a similar method to my answer to your previous question. That is, you can use rook theory.

Given a subset (or "board") $B \subseteq [n] \times [n]$, let $r_{B,k}$ be the the $k$-th rook number, that is, the number of placements of $k$ rooks on the board $B$ so that no two rooks are in the same row or column. Let $h_{B,k}$ be the $k$-th hitting number of $B$, defined to be the number of permutations $\sigma \in S_n$ so that $\{(i,j) \in B | \sigma(i) = j\} = k$. Put another way - we call any $1$ on the adjacency matrix of $\sigma$ that lands on the board $B$ a hit of $\sigma$. Then $h_{B,k}$ is the number of permutations $\sigma \in S_n$ with exactly $k$ hits in $B$.

Then the following relation holds:

\begin{equation} \sum_{k=0}^n h_{B,k} x^k = \sum_{k=0}^n r_{B, k} (n-k)! (x-1)^k. \tag{*} \end{equation}

See, e.g., Theorem 1 in Remmel's notes here. This equation (*) allows you to find hit numbers from rook numbers, and vice verse.

Using the same notation there, let $B = B_1 \oplus \cdots \oplus B_k$ where $B_i = [n_i] \times [n_i]$. That is, $B \subseteq [n] \times [n]$ is the block-diagonal set consisting k disjoint squares with dimensions $n_i \times n_i$. Then the answer to your question is the $k$-th hit number $h_{B,k}$ of the board $B$. Thus it remains to find the rook numbers $r_{B,k}$; then we can use (*) to find $h_{B,k}$.

Define the rook polynomial $r_B(x)$ of a board $B \subseteq [n] \times [n]$ to be $$r_B(x) = \sum_{k=0}^n r_{B,k} x^k.$$ This is slightly different, but equivalent to, the definition of $r_B(x)$ I gave in the previous answer. But we still have $$r_{B_1}(x) r_{B_2}(x) = r_{B_1 \oplus B_2}(x).$$

Then if $B$ is the full square $[n] \times [n]$, we have $$r_B(x) = \sum_{k=0}^n {n \choose k}^2 \, k!\, x^k.$$ Call this $L_n(x)$. Then to find the partial derangement numbers, expand $r_B(x) = L_{n_1}(x) \cdot \cdots \cdot L_{n_k}(x)$ and apply (*).

Example: Let $n=3$, with $n_1 = 1$, $n_2 = 2$. Compute $L_{1}(x) = 1+x$, $L_2(x) = 1 + 4x + 2x^2$. Then if $B$ is the block diagonal subset $[1] \times[1] \oplus [2] \times [2]$ Then $$r_B(x) = L_1(x) L_2(x) = 1 + 5x + 6x^2 + 2x^3.$$ Send each power $x^k$ to $(n-k)! (x-1)^k$ to get

\begin{align*}3! + 5\cdot 2! (x-1) + 6 \cdot 1! (x-1)^2 + 2 \cdot 0!(x-1)^3 &= 4x + 2x^3\\ &= \sum_{k = 0}^n h_{B,k} x^k.\end{align*}

This means that if $B$ that the number of permutations $\sigma \in S_n$ with $1$ hits in $B$ is $4$, the number of permutations $\sigma \in S_n$ with $3$ hit in $B$ is 2, and there are no permutations with $0$ or $2$ hits. (note that the coefficients here sum to $2 + 4 = 6 = 3!$, the number of permutations of $S_3$.)

To verify, suppose $1$ is blue and $2,3$ are red; we have the permutations

\begin{align*} 123 \rightarrow 123\,\, \text{(3 hits)} \\ 123 \rightarrow 132\,\, \text{(3 hits)} \\ 123 \rightarrow 213\,\, \text{(1 hit)}\hphantom{1} \\ 123 \rightarrow 231\,\, \text{(1 hit)}\hphantom{1} \\ 123 \rightarrow 312\,\, \text{(1 hit)}\hphantom{1} \\ 123 \rightarrow 321\,\, \text{(1 hit)}\hphantom{1} \\ \end{align*}