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How many arrangements of the numbers $1,1,1,2,2,3,3,3,4,4$ are there in which the adjacent numbers are not equal?

I believe that the total number of possible arrangements is $\frac{10!}{ 3! \cdot 2! \cdot 3! \cdot 2!}$, however, I do not understand what else I need to calculate so that I can subtract it from the total.

I believe that this is called the inclusion-exclusion principle

N. F. Taussig
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DeviceGucci
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  • Inclusion-exclusion can be made to work here, but it might not be the best approach. I'd be inclined to try list the ways to position three $1$'s non-contiguously (breaking the string of ten positions into $2,3$ or $4$ smaller groups, where it is easier to place digits non-consecutively). It is a good practice to give a complete problem statement in the body of your Question, so that it is self-contained (independent of the title). – hardmath Nov 12 '21 at 03:56
  • Sorry, it's my first time using this, but the title is the entire problem statement, so I thought that it was fine. – DeviceGucci Nov 12 '21 at 04:05
  • I went ahead and made the minimal edit to have that in the body of your Question. Of course you can rollback the edit, such as if it changes your meaning in some way. – hardmath Nov 12 '21 at 04:12
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    @hardmath i put an another method different from P.I.E if you still interested in – Not a Salmon Fish Nov 12 '21 at 12:06

2 Answers2

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To find what to subtract, use the inclusion-exclusion. That is do something like this: If the first two digits are $1$'s, the number of arrangements is compute it, add the number of arrangements where the first two digits are $2$'s, $3$s or $4$s. Repeat with the number of arrangements where the second and the third digits are the same, etc. Then you would have to subtract the numbers of arrangements that belong to two of these sets. You will not need to also count triple intersections because those are fortunately empty.

markvs
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It is said that no consecutive same number at all.Then , if we talk about $1's$ , then $111$ or $11$ is forbidden ,right ? Well, it is suggested to use inclusion-exclusion method , it is okey but a little cumbersome . if you want a shortcut , i will recommend you a very powerful method.You can easily solve any question in this type with this method.

When we pause on your question again , it is brefily asked that how many arrangements there are where there is not any consecutive same number.So , we are looking for the cases where all same numbers are separate.

To solve this messy process , i will use analytic combinatorics techniques. In analytic combinatorics , we call such arrangements as Smirnov or Carlitz words.I am putting a link for you link of the suggested book You can find detaily explanation here about my method , but i will not get in explanation .

Before , i start my solution to prevent confusions in my method. I want to name the numbers as letters such that $1,2,3,4$ will be represented by $A,B,C,D$ , respectively. So , our question is the same as how many arrangements are there where no same letter are adjacent. Then , lets start..

A generating function for the number of Smirnov words over an n-ary alphabet is given by $$\begin{align*} \left(1-\frac{nz}{1+z}\right)^{-1} \end{align*}$$

Here we consider an alphabet $\mathcal{K}=\{A,B,C,D\}$ with $n=4$ letters. Using $[z^k]$ to denote the coefficient of $z^k$ of a series we calculate

\begin{align*} &\color{green}{[A^3B^2C^3D^2]\left(1-\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^{-1}}\\ &\qquad=[A^3B^2C^3D^2]\sum_{j=0}^{\infty}\left(\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^j\\ &\qquad=[A^3B^2C^3D^2]\sum_{j=4}^{10}\left(\sum_{a\in\mathcal{V}}\frac{a}{1+a}\right)^j\\ &\qquad=[A^3B^2C^3D^2]\sum_{j=4}^{10}\left((A-A^2 +A^3 )+\left(B-B^2 \right)\right.\\ &\qquad\qquad\qquad\qquad\qquad\qquad\left.+(C-C^2 +C^3 )+(D-D^2 )\right)^j\\ &\qquad=24- 360+2580-10920+28000-40320+25200\\ &\,\,\color{red}{\qquad=4,204} \end{align*}

I am putting a link for calculating generating functions CALCULATION LINK

Not a Salmon Fish
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