Partial Derangement formula for permutation with repeated elements 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)? 
 A: 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*}
