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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 $26^6.)$. $\endgroup$ – Thomas Andrews Mar 31 '15 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. $\endgroup$ – Marko Riedel Mar 31 '15 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 Mar 31 '15 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. $\endgroup$ – Marko Riedel Mar 31 '15 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 Mar 31 '15 at 2:09
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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 Mar 31 '15 at 3:11
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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 Mar 31 '15 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 Mar 31 '15 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 Mar 31 '15 at 4:07

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