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In how many ways can we assign the numbers $\{ 1,2,\dots,s\}$ to the variables $a_1, \dots,a_k$ such that each number appears at least once, and exactly $t$ of the pairs $(a_1,a_2), (a_2,a_3), \dots,(a_k,a_1)$ are identical?

For instance, let $k = 6$, $s=3$, and $t = 2$. Then the following are some of the valid sequences:

$123321$, $212322$, $333121$, $113233$, etc.

I tried first choosing the $t$ identical pairs (in $\binom{k}{t}$ ways), "packaging" them into a single element. Then the problem reduces to finding the number of ways $s$ values can be assigned such that each variable appears at least once, and no same value appears consecutively. This is where I got stuck, any help/pointers would be appreciated.

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It is very likely that an analysis with cases and binomial coefficients, as you suggested, will eventually work out. However, there is a general technique, called the Goulden-Jackson cluster method of counting strings that avoid particular patterns that solves most problems of this type in one fell stroke. This method has several advantages.

  1. It is no harder to learn than solving your original problem.
  2. It efficiently produces a generating function for the enumeration problem.
  3. A Maple implementation of the method is freely available, thanks to Doron Zeilberger and John Noonan.

I highly recommend spending an hour or two learning the method. If so, the analysis would go like this.

Let $a_{k,s,t}$ be the number of sequences of length $k$ composed of digits $1,\ldots, s$ with exactly $t$ pairs of consecutive digits being identical. Define the generating function $F_s(z,w)=\sum_{k,t}a_{k,s,t}\,z^k w^t.$ The GJ method then gives $$F_s(z,w)=\frac1{1-sz-C},$$ where $C$ is the weight enumerator of the so-called clusters. The method also gives $$C=-\frac{sz^2(w-1)}{1+z-zw},$$ resulting in $$F_s(z,w)=\frac{1+z-zw}{1-(s-1)z-zw}.$$ Because of the basic form of this generating function, it is straightforward to extract its coefficients. Then, the solution to your problem is $$a_{k,s,t}=[z^kw^t]F_s(z,w)= s\binom{k-1}{t}(s-1)^{k-1-t}.$$

Note: The exact problem you stated had a couple conditions that I did not include.

  1. Each digit must appear at least once.
  2. A pair of identical digits can "circle back" from the end of the sequence to the beginning.

The first item is easy to fix, just with subtraction: the number of such sequences with each digit appearing is just $a_{k,s,t}-a_{k,s-1,t}$. The second item is more subtle, but there is an adaption of the GJ method for cyclic words.