A bag contains $2$ red, $2$ green, $2$ black, $3$ white, $1$ yellow, $1$ brown and $1$ orange ball.
In how many ways the balls be arranged in a row by taking all of them together if balls of same color cannot be placed adjacent?
I am aware that $\dfrac{12!}{2!\times 2! \times 2! \times 3!}$ arrangements are possible. But these arrangements contain balls of same colors placed adjacent also. How to solve this problem?
This answer is based upon a generating function of generalized Laguerre polynomials \begin{align*} L_k^{(\alpha)}(t)=\sum_{i=0}^k(-1)^k\binom{k+\alpha}{k-i}\frac{t^i}{i!}\tag{1} \end{align*}
The Laguerre polynomials have some remarkable combinatorial properties and one of them is precisely suited to answer problems of this kind. This is nicely presented in Counting words with Laguerre series by Jair Taylor.
We encode the colors (r)ed, (g)reen, (b)lack, (w)hite, (y)ellow, (B)rown, (o)range of the balls with the letters \begin{align*} \{r,g,b,w,y,B,o\} \end{align*} and are looking for words of length $12$ built from \begin{align*} r,r,g,g,b,b,w,w,w,y,B,o \end{align*} which have the property that they contain no consecutive equal letters. These words are called Carlitz words or Smirnov words.
We find in section 2 of the referred paper Laguerre polynomials $l_k(t)$ defined by their generating function \begin{align*} \sum_{k=0}^\infty l_k(t)x^k=e^{\frac{tx}{1+x}} \end{align*} The first few such polynomials are \begin{align*} l_0(t)&=1\\ l_1(t)&=t\\ l_2(t)&=\frac{1}{2}t^2-t\\ l_3(t)&=\frac{1}{6}t^3-t^2+t\tag{2} \end{align*} These polynomials are a specific form of Laguerre polynomials (1), namely $$l_k(t)=(-1)^kL_k^{(-1)}(t)$$
Theorem 2.1 in the referred paper states: Given nonnegative integers $n_1,\ldots,n_k$, the number of $k$-ary Carlitz words with the letter $i$ used exactly $n_i$ times is \begin{align*} \int_{0}^\infty e^{-t}\left(\prod_{i=1}^kl_{n_i}(t)\right)\,dt\tag{3} \end{align*}
Since we have three characters $r,g,b$ each twice, one character $w$ three times and three characters $y,B,o$ each occurring once, we set \begin{align*} &n_1=n_2=n_3=2,\\ &n_4=3,\\ &n_5=n_6=n_7=1 \end{align*}
We apply theorem 2.1. and obtain using (2) and (3) and with some help of Wolfram Alpha \begin{align*} \int_{0}^\infty&e^{-t}\left(\prod_{i=1}^7l_{n_i}(t)\right)\,dt\\ &=\int_{0}^\infty e^{-t}\left(l_2(t)\right)^3l_3(t)\left(l_1(t)\right)^3\,dt\\ &=\int_{0}^\infty e^{-t}\left(\frac{1}{2}t^2-t\right)^3\left(\frac{1}{6}t^3-t^2+t\right)t^3\,dt\\ &=\int_{0}^\infty e^{-t}\left(\frac{1}{48}t^{12}-\frac{1}{4}t^{11}+\frac{9}{8}t^{10} -\frac{29}{12}t^9+\frac{5}{2}t^8-t^7\right)\,dt\\ &=3301200 \end{align*}
This question is equivalent to
"In how many different ways can we arrange letters of the word AABBCCDDDEFG such that no two adjacent letters are alike?"
I have been doing many such questions in past. This tool was very useful. Following is the result obtained from this tool.
[see generated question $24$. Thought of retying it here. but as i said in my comment, it is very long and pasting the screenshot here]
