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How many strings are there of seven lowercase letters that have the substring tr in them?

So I am having a little problem with this question, I know that the total number of combinations is $26^6$ but there is double counting on some of the combinations.

For example, when you have a case that contains multiples 'tr' then it will be counted multiple times depending on the location of tr, even though its the same string.

t r t r t r a

Any advice on what to subtract to remove this double counting?

Thanks for any help!

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    $\begingroup$ Better to count the number of ways to not get an instance of "tr," and then subtract that from $26^7$. (The total strings is $26^7$, not $26^6.)$. $\endgroup$ Commented Mar 31, 2015 at 1:25
  • $\begingroup$ I suggest two recurrences for strings not containing TR ending in T and not ending in T, solving these, and compute the value for length n=7. $\endgroup$ Commented Mar 31, 2015 at 1:30
  • $\begingroup$ But 'tr' is a substring that must stay together, meaning you can consider it as one object. So instead of having 7 spaces to fill you are only filling 5 then the substring takes up the other two as one making 6. Or did i do that totally wrong? $\endgroup$
    – Anon
    Commented Mar 31, 2015 at 1:30
  • $\begingroup$ Because these parameters are very reasonable inclusion-exclusion will also work here. Apply stars-and-bars when you count the configurations containg $q$ copies of the string TR. $\endgroup$ Commented Mar 31, 2015 at 1:55
  • $\begingroup$ Never used stars-and-bars before but something like this? ★ |★ ★ ★ ★ ★ - ★ |★ |★ ★ ★ ★ - ★ ★ |★ |★ ★ ★ - ★ ★ ★ |★ |★ ★ - ★ ★ ★ ★ |★ |★ - ★ ★ ★ ★ ★ |★ So there are 6 different spots the 'tr' can go and then the other spots should not contain 'tr' so remove 2 letters from the 5 spots. Then the final equation would be 26^6 - 6 x 2^5? $\endgroup$
    – Anon
    Commented Mar 31, 2015 at 2:09

2 Answers 2

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Let $a_n$ be the number of strings of length $n$ with no tr's. An $(n-1)$-long string can be extended in $26$ ways unless it ends in "t", in which case it can only be extended in $25$ ways. So $$a_n= 26a_{n-1}-a_{n-2}$$ for $n\ge2$; the initial conditions are $a_0=1$ and $a_1=26$.

You can give a formula for $a_n$, but to compute $a_7$ it suffices to run out the recurrence. The final answer is $26^7-a_7 = 71,112,600$.

In case there's doubt, another way to verify the recurrence is to write down the $26\times26$ transition matrix for letters. All of the entries are $1$'s except for a single $0$ off the diagonal (corresponding to the prohibited tr). The characteristic polynomial is computed to be $z^{24}(z^2-26z+1)$.

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  • $\begingroup$ Thank you this helped a ton! $\endgroup$
    – Anon
    Commented Mar 31, 2015 at 3:11
  • $\begingroup$ Does it extend to string of length more than 2? I'm working on a similar problem, but the substring is arbitrary. I write some script and found that $a_n = 26a_{n-1} - a_{n-2} - a_{n-3} - \ldots$ doesn't work. $\endgroup$ Commented Aug 6, 2021 at 15:47
  • $\begingroup$ Ok I get it now. Should be $a_n = 26a_{n-1} - a_{n-ss}$, where $ss$ is the length of the substring. $\endgroup$ Commented Aug 6, 2021 at 21:27
  • $\begingroup$ For general substrings, a good approach is to let $a_{n,k}$ be the number of strings of length n ending in the k-long prefix of the given substring, then write down a recurrence (the answer depends on how the substring overlaps with shifts of itself, so different substrings even of the same length can give different answers.) $\endgroup$
    – Tad
    Commented Aug 8, 2021 at 2:03
  • $\begingroup$ @Tad Yes I've just bumped into the overlap! Is there any intuition for the fact that the number of strings containing tt is less than the number of strings containing tr? $\endgroup$ Commented Aug 9, 2021 at 22:02
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There are two possible solutions here, one using recurrences and the other one using PIE.

Recurrence.

The recurrences use two sequences $\{a_n\}$ and $\{b_n\}$ which count strings not containing the two-character pattern that end in the first character of the pattern and that do not. This gives

$$a_1 = 1 \quad\text{and}\quad b_1 = 25$$ and for $n\gt 1$ $$a_n = a_{n-1} + b_{n-1} \quad\text{and}\quad b_n = 24a_{n-1} + 25b_{n-1}.$$

These recurrences produce for $1\le n\le 7$ the sequence of sums $\{a_n+b_n\}$ which is $$26, 675, 17524, 454949, 11811150, 306634951, 7960697576,\ldots$$ so that the answer to the problem is (count of no ocurrences) $$7960697576.$$

Inclusion-Exclusion.

Let $M_{\ge q}$ be the set of strings containing at least $q$ ocurrences of the two-character pattern and let $M_{=q}$ be the set containing exactly $q$ ocurrences of the pattern. Then by inclusion-exclusion we have

$$|M_{=0}| = \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k |M_{\ge k}|.$$

Note however that $$|M_{\ge k}| = 26^{n-2k} {n-2k + k\choose k} = 26^{n-2k} {n-k\choose k}.$$

This is because when we have $k$ copies of the pattern there are $n-2k$ freely choosable letters that remain. Hence we have a total of $n-2k+k=n-k$ items to permute. We then choose the $k$ locations of the patterns among the $n-k$ items.

This gives the formula $$|M_{=0}| = \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k \times 26^{n-2k} \times {n-k\choose k}.$$

a :=
proc(n)
    option remember;

    if n=1 then return 1 fi;

    a(n-1)+b(n-1);
end;


b :=
proc(n)
    option remember;

    if n=1 then return 25 fi;

    24*a(n-1)+25*b(n-1);
end;

ex_pie :=
proc(n)
    option remember;

    add((-1)^q*26^(n-2*q)*binomial(n-q,q),
        q=0..floor(n/2));
end;

Proof that the two answers are the same.

Introduce the generating functions $$A(z) = \sum_{n\ge 0} a_n z^n \quad\text{and}\quad B(z) = \sum_{n\ge 0} b_n z^n.$$

Observe that the correct intial value pair is $a_0 = 0$ and $b_0 = 1.$

Multiply the two recurrences by $z^n$ and sum over $n\ge 1$ to get $$A(z) - 0 = z A(z) + z B(z) \quad\text{and}\quad B(z) - 1 = 24 z A(z) + 25 z B(z).$$

Solve these two obtain $$A(z) = \frac{z}{z^2-26z+1} \quad\text{and}\quad B(z) = \frac{1-z}{z^2-26z+1}.$$

This yields the following generating function $G(z)$ for $\{a_n+b_n\}:$ $$G(z) = \frac{1}{z^2-26z+1}.$$

On the other hand we have $$G(z) = \sum_{n\ge 0} z^n \left(\sum_{k=0}^{\lfloor n/2\rfloor} 26^{n-2k} (-1)^k {n-k\choose k}\right).$$

This is $$\sum_{k\ge 0} 26^{-2k} (-1)^k \sum_{n\ge 2k} 26^n z^n {n-k\choose k} \\ = \sum_{k\ge 0} 26^{-2k} (-1)^k \sum_{n\ge 0} 26^{n+2k} z^{n+2k} {n+2k-k\choose k} \\ = \sum_{k\ge 0} z^{2k} (-1)^k \sum_{n\ge 0} 26^n z^{n} {n+k\choose k} = \sum_{k\ge 0} z^{2k} (-1)^k \frac{1}{(1-26z)^{k+1}} \\ = \frac{1}{1-26z} \sum_{k\ge 0} z^{2k} (-1)^k \frac{1}{(1-26z)^{k}} = \frac{1}{1-26z} \frac{1}{1+z^2/(1-26z)} \\ = \frac{1}{1-26z+z^2}.$$

This establishes the equality of the generating functions which was to be shown.

Closed form and OEIS entry.

The roots of the denominator of the generating function are $$\rho_{1,2} = 13 \pm 2\sqrt{42}.$$ Writing $$\frac{1}{1-26z+z^2} = \frac{1}{(z-\rho_1)(z-\rho_2)} = \frac{1}{\rho_1-\rho_2} \left(\frac{1}{z-\rho_1}-\frac{1}{z-\rho_2}\right) \\ = \frac{1}{4\sqrt{42}} \left(\frac{1}{\rho_1}\frac{1}{z/\rho_1-1} -\frac{1}{\rho_2}\frac{1}{z/\rho_2-1}\right) \\ = \frac{1}{4\sqrt{42}} \left(-\frac{1}{\rho_1}\frac{1}{1-z/\rho_1} +\frac{1}{\rho_2}\frac{1}{1-z/\rho_2}\right).$$

We now extract coefficients to get $$[z^n] G(z) = \frac{1}{4\sqrt{42}} \left(\rho_2^{-n-1}-\rho_1^{-n-1}\right).$$

Since $\rho_1\rho_2 = 1$ this finally becomes $$[z^n] G(z) = \frac{1}{4\sqrt{42}} \left(\rho_1^{n+1}-\rho_2^{n+1}\right)$$

which is the sequence $$26, 675, 17524, 454949, 11811150, 306634951, 7960697576, \\ 206671502025, 5365498355074, 139296285729899, \ldots$$

This is OEIS A097309 which has additional material and where in fact we find a copy of the problem statement that initiated this thread.

Alternative derivation of the closed form of $G(z).$

This uses the following integral representation. $${n-k\choose k} = \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^{n-k}}{w^{k+1}} \; dw.$$

This gives for the inner sum $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^n}{w} \left(\sum_{k=0}^{\lfloor n/2\rfloor} 26^{n-2k} (-1)^k \frac{1}{(1+w)^k w^k}\right) \; dw.$$

Note that the defining integral is zero when $\lfloor n/2\rfloor \lt k \le n,$ so this is in fact $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^n}{w} \left(\sum_{k=0}^n 26^{n-2k} (-1)^k \frac{1}{(1+w)^k w^k}\right) \; dw.$$

Simplifying we obtain $$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(1+w)^n}{w} 26^n \frac{(-1)^{n+1}/26^{2(n+1)}/(1+w)^{n+1}/w^{n+1}-1} {(-1)/26^2/(1+w)/w-1} \; dw$$ or $$\frac{1}{2\pi i} \int_{|w|=\epsilon} (1+w)^{n+1} 26^n \frac{(-1)^{n+1}/26^{2(n+1)}/(1+w)^{n+1}/w^{n+1}-1} {(-1)/26^2-w(w+1)} \; dw$$

The difference from the geometric series contributes two terms, the second of which has no poles inside the contour, leaving just

$$\frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{(-1)^{n+1}}{26^{n+2}} \frac{1}{w^{n+1}} \frac{1}{(-1)/26^2-w(w+1)} \; dw.$$

It follows that $$G(z) = \sum_{n\ge 0} z^n [w^n] \frac{(-1)^{n+1}}{26^{n+2}} \frac{1}{(-1)/26^2-w(w+1)}.$$

What we have here is an annihilated coefficient extractor which simplifies as follows. $$\frac{-1}{26^2} \sum_{n\ge 0} (-z/26)^n [w^n] \frac{1}{(-1)/26^2-w(w+1)} \\ = \frac{-1}{26^2} \frac{1}{(-1)/26^2+z/26(-z/26+1)} = -\frac{1}{-1+z(-z+26)} \\ = \frac{1}{1-26z+z^2}.$$

This concludes the argument.

There is another annihilated coefficient extractor at this MSE link.

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  • $\begingroup$ This is amazing, but I am confused because for a string of length 1 there should be 0 strings that contain 'tr' then strings of length 2 should only contain 1 ('tr') then Strings of length 3 should be 52 i believe (t,r,26)or(26,t,r) but you came up with 26,675,172554? $\endgroup$
    – Anon
    Commented Mar 31, 2015 at 3:50
  • $\begingroup$ Oh wow never mind that is how many strings that don't contain the 'tr' so for 7 characters 26^7 - 7960697576 = 71,112,600. $\endgroup$
    – Anon
    Commented Mar 31, 2015 at 3:52
  • $\begingroup$ Seriously thank you so much for taking all this time to solve this, you helped me actually understand where the numbers were coming from. $\endgroup$
    – Anon
    Commented Mar 31, 2015 at 4:07

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