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There are 7 5 = 16,807 strings of length 5 formed from the 7 letters a, b, c, d, e, f, & g.   For example, one such string is ebdbg.
In how many of these 16,807 strings are the letters b & d adjacent (appearing in either order  bd  or  db ) ?
In other words, how many of these 16,807 strings contain either bd as a substring or contain db as a substring?
user3134725
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  • Do you include a string without both $b$ and $d$ as one of the required strings? It's unclear what you are counting. If you are counting the strings such that every occurrence of $b$ is accompanied by a neighboring $d$, then a string like $aaaaa$ is one of them. But your question may also mean that you are counting a string with at least one $b$. – Batominovski Oct 05 '18 at 20:01
  • The string aaaaa contains neither bd nor db as a substring, and so the string aaaaa is not counted. – user3134725 Oct 05 '18 at 20:02
  • Do you include a string with multiple b's and d's as one of the desired strings even if one of the occurrences of b is not adjacent to a d or vice versa? For example, do you count bdaab? How about bdbaa? – JMoravitz Oct 05 '18 at 20:03
  • Well, it is not clear from the question as the statement that $b$ and $d$ are adjacent in the string $aaaaa$ is also a true statement, if you write the statement like this: $$\forall x, \text{ if }x\text{ appears in the string }aaaaa\text{ and }x=b,\text{ then }\exists y, y\text{ is a neighbor of }x\text{ and }y=d.$$ – Batominovski Oct 05 '18 at 20:06
  • Include both bdaab and bdbaa in the count since both bdaab and bdbaa contain either bd or db as a substring. – user3134725 Oct 05 '18 at 20:06
  • Why do you include $bdaab$? The last $b$ is not adjacent to any $d$. – Batominovski Oct 05 '18 at 20:07
  • Is your question actually this? Find the number of strings of length $5$ composed by letters $a,b,c,d,e,f,g$ such that there exists a substring of the form $bd$ or $db$. – Batominovski Oct 05 '18 at 20:08

2 Answers2

1

For a nonnegative integer $n$, a string of length $n$ composed by letters $a,b,c,d,e,f,g$ containing $bd$ or $db$ as a substring is said to be $n$-suitable. Let $x_n$ denote the number of $n$-suitable strings. Clearly, $x_0=0$, $x_1=0$, $x_2=2$, and $x_3=26$.

For $n>1$, it can be seen that $$x_n=5x_{n-1}+10x_{n-2}+10x_{n-3}+\ldots+10x_1+2\cdot 7^{n-2}+2\cdot 7^{n-3}+\ldots +2\cdot 7^0\,.\tag{*}$$ To show this, look at the end of an $n$-suitable string $t_1t_2\ldots t_n$. If $t_n\notin\{b,d\}$, then the string $t_1t_2\ldots t_{n-1}$ must be $(n-1)$-suitable. There are $5$ choices for $t_n$, and so this gives $5x_{n-1}$ possible $n$-suitable strings with $t_n\notin\{b,d\}$.

Now, suppose that $t_n\in\{b,d\}$. Let $k$ be the smallest positive integer such that $$t_k=t_{k+1}=\ldots=t_{n}\,.$$ Clearly, $k>1$. If $t_{k-1}\in\{b,d\}\setminus\{t_n\}$, then $t_1t_2\ldots t_{k-2}$ can be arbitrary, so we have $2\cdot 7^{k-2}$ possibilities, recalling that there are two choices for $t_n$.

If $t_{k-1}\notin\{b,d\}$, then $t_1t_2\ldots t_{k-2}$ is a $(k-2)$-suitable string. There are $5$ choices of $t_{k-1}$ and $2$ choices for $t_n$. Thus, we have in total $10x_{k-2}$ possible varieties.

From (*), we have $$x_{n-1}=5x_{n-2}+10x_{n-3}+\ldots+10x_1+2\cdot 7^{n-3}+2\cdot 7^{n-4}+\ldots+2\cdot7^0$$ for $n>2$. Subtracting the above result from (*), we get $$x_n-x_{n-1}=5x_{n-1}+5x_{n-2}+2\cdot 7^{n-2}\,,$$ so $$x_n-6x_{n-1}-5x_{n-2}=2\cdot7^{n-2}\,.$$ Solving the recurrence relation gives $$x_n=7^n-\left(\frac{7-2\sqrt{14}}{14}\right)(3-\sqrt{14})^n-\left(\frac{7+2\sqrt{14}}{14}\right)(3+\sqrt{14})^n\,.$$

However, to get $x_5$, you do not need the general formula. Just keep using the recurrence relation, you will get your desired answer. In other words, you compute $x_4$ to get $$x_4=6x_3+5x_2+2\cdot 49=264\,.$$ Then, $$x_5=6x_4+5x_3+2\cdot 343=2400\,.$$

Batominovski
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1

This answer is based upon the Goulden-Jackson Cluster Method which is a convenient technique to derive a generating function for problems of this kind.

We consider words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{a,b,c,d,e,f,g\}$$ and the set $B=\{bd,db\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $f(s)$ with the coefficient of $s^n$ being the number of searched words of length $n$.The wanted number of words of length $5$ which do contain one of the bad words is consequently \begin{align*} \color{blue}{7^5-[s^5]f(s)} \end{align*} with $[s^n]$ denoting the coefficient of $f(s)$ of a series.

According to the paper (p.7) the generating function $f(s)$ is \begin{align*} f(s)=\frac{1}{1-ds-\text{weight}(\mathcal{C})} \end{align*} with $d=|\mathcal{V}|=7$, the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with \begin{align*} \text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[bd])+\text{weight}(\mathcal{C}[db]) \end{align*}

We calculate according to the paper \begin{align*} \text{weight}(\mathcal{C}[bd])&=-s^2-\text{weight}(\mathcal{C}[db])s\\ \text{weight}(\mathcal{C}[db])&=-s^2-\text{weight}(\mathcal{C}[bd])s\\ \end{align*} and get \begin{align*} \text{weight}(\mathcal{C}[bd])&=\text{weight}(\mathcal{C}[db])=\frac{-s^2}{1+s}\\ \text{weight}(\mathcal{C})&=\text{weight}(\mathcal{C}[bd])+\text{weight}(\mathcal{C}[db])\\ &=\frac{-2s^2}{1+s} \end{align*}

We can now calculate $f(s)$

\begin{align*} \color{blue}{f(s)}&=\frac{1}{1-ds-\text{weight}(\mathcal{C})}\\ &=\frac{1}{1-7s+\frac{2s^2}{1+s}}\\ &=\frac{1+s}{1-6s-5s^2}\\ &=1+7s+47s^2+317s^3+2\,137s^4+\color{blue}{14\,407}s^5+97\,127s^6+\cdots\tag{1} \end{align*} where the last line was done with some help of Wolfram Alpha.

We finally conclude from (1) the number of words containing at least one substring from $\{bd,db\}$ is \begin{align*} 7^5-[x^5]f(s)=16\,807-14\,407=\color{blue}{2\,400} \end{align*}

Hint: Some aspects of this approach are provided in the Add-on part of this answer.

Markus Scheuer
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