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What is the formula for doing this type of problems?

N. F. Taussig
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Joshua Leung
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  • The idea is to go recursively : which strings of length $2$ don't contain $10,21$ or $20$ as a substring? Now, how do you go from a string of length $2$ to a string of length $3$? By adding one digit, where you have to be careful. Now from length $3$ to length $4$? – Sarvesh Ravichandran Iyer Sep 11 '19 at 13:10
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    There is an easy way of doing this particular problem if you see it right. The idea of a general formula seems less useful here because it might stop you from seeing the easy answer. – Mark Bennet Sep 11 '19 at 13:12
  • @MarkBennet Very true. It's kinda frustrating to me how often people want some formula for some specific situation. Even if it exists, chances are that it is either not closed, or it's not intuitive in any way. I much prefer understanding and solving a problem as is. – Rushabh Mehta Sep 11 '19 at 13:14

2 Answers2

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Here we use PIE the inclusion-exclusion principle to count the number of valid $4$-letter words from the alphabet $\{0,1,2\}$ which do not contain the bad words $\{10,20,21\}$.

In order to do the job some kind of bookkeeping is helpful. We consider \begin{align*} .\ .\ .\ . &-\left(10\ .\ .|20\ .\ .|21\ .\ .\right)\tag{1}\\ &+\left(10\ 10|10\ 20|10\ 21|20\ 20|20\ 21|210\ .|21\ 21\right)\tag{2}\\ \end{align*}

Comment:

  • In (1) we count all $4$-letter words indicated by four dots which gives $3^4$. Then we subtract all words which contain at least one bad word.

    Since $10$ consumes two characters and two are left for free assignment, we count $\binom{3}{1}3^2$ words of this kind and similarly in the other cases with the bad words $20$ and $21$.

  • In (2) we add words containing two bad words as compensation for those which we've subtracted twice in (1), noting that we also have to consider overlaps $21$ with $10$ giving $\color{blue}{210}$.

  • No more cases are left to consider, since words containing three or more bad words have length $>4$.

We obtain according to (1) to (2): \begin{align*} 3^4&-\left(\binom{3}{1}3^2+\binom{3}{1}3^1+\binom{3}{1}3^1\right)\\ &\quad+\left(\binom{2}{2}3^0+2\binom{2}{2}3^0+2\binom{2}{2}3^0+\binom{2}{2}3^0+2\binom{2}{2}3^0+\binom{2}{1}3^1 +\binom{2}{2}3^0\right)\\ &=81-(27+27+27)+(1+2+2+1+2+6+1)\\ &=81-81+15\\ &\,\,\color{blue}{=15} \end{align*}

Markus Scheuer
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Hint. The first letter is $0$, $1$ or $2$. Apart from the last letter, the letters should obey the following constraints: every $1$ is followed by $1$ or $2$ and every $2$ is followed by $2$. Just draw by hand the corresponding graph (actually a forest) and you will find the answer: 15.

J.-E. Pin
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