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Got a combinatorics question and honestly no idea how to get by it. Would love some reference to material related to the topic as well as my course notes do a great job of not having any examples.

A change in a binary string is an occurrence of two consecutive terms in the string that are different (that is, one is a $0$ and the other is a $1$). For example, in the binary string $1001$, there are two changes: the $10$ at the beginning and the $01$ at the end. In $1010$, there's $3$ changes because: $10$ at the beginning, $01$ in the middle, $10$ at the end.

How many binary strings of length $n$ have exactly $k$ changes? Where a change is the existence of $10$ or $01$s

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  • $\begingroup$ Recursion often works well for problems like this. If you know the answer for strings of length $n-1$ (and all $k$) can you answer the question for strings of length $n?$ $\endgroup$ – saulspatz Sep 13 '18 at 16:02
  • $\begingroup$ While I can see the logic I'm not sure how to actually go about proving it through anything other than inspection. This is a new topic for me and I don't have any experience writing formal proofs for this. They also mention that the answer should be in forms of binomial coefficients. $\endgroup$ – Bone Sep 13 '18 at 16:07
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Yes, your approach is correct. Along the string, the digits change $k$ times. Such changes can be placed between the $i$th digit and $(i+1)th$ digit for $i=1,\dots, n-1$. This can be done in $\binom{n-1}{k}$ ways. Finally note that each string starts with a $0$ or $1$ so the total number of such strings is $2\binom{n-1}{k}$.

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  • $\begingroup$ Sound logic. Just to confirm, the question uses the phrase 'Justify your answer'. Is this sufficient justification or is there a more formal way of proving this somehow? $\endgroup$ – Bone Sep 13 '18 at 16:10
  • $\begingroup$ @Bone I gave you a brief response which, in my opinion, is sufficiently formal. $\endgroup$ – Robert Z Sep 13 '18 at 16:15
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automaton with counters

Let $A$ be binary strings that ends with $0$ and $B$ that ends with $1$.

$x$ is a counter variable for the length of an accepted word, while $t$ will count the transitions. Then:

$A = xA + txB + x$

$B = xB + txA + x$ and

$A+B = {2x \over 1-x-tx} = 2x + 2x^2(1+t) + 2x^3(1+t)^2 +...$

The required number for length $n$ and $k$ changes is the coefficient of $x^nt^k$.

For example, if $n = 4$ and $k = 2$ we have six good words: 0100, 0110, 0010, 1011, 1001, and 1101.

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