Permutations of MISSISSIPPI not containing adjacent identical letters

Question

In how many arrangements of the letters of the word $MISSISSIPPI$ are no two identical letters stand next to each other?

We can use Inclusion Exclusion principle but there are too many number of sets to consider. For example, we need to count those permutations in which $(SSSS)$, $((SSS), (S))$ occur but not together, $((SS), (SS))$, occur but not together etc. and similarly for $I$s. The problem is from Counting and Configurations by Jiri Herman, Radan Kucera and Jaromir Simsa. The book derives a complicated expression similar to Mobius Inversion formula and uses it to obtain the answer. Is there a simpler way? The answer to the current question is 2016.

There is actually an explicit formula for this sort of problem involving Laguerre polynomials. See here — Jair Taylor, Oct 19 '18 at 03:43
@JairTaylor You mentioned in that post in 2012 that "...has not been published anywhere...not sure what to expose it publicly until it is proven..." So have you published it somewhere like in MAA journals? — Lee David Chung Lin, Oct 19 '18 at 14:56
@LeeDavidChungLin Oh, yeah, eventually I published it here. But the result was originally due to Ira Gessel. — Jair Taylor, Oct 19 '18 at 15:04

N. Shales · Accepted Answer · 2018-10-19T03:32:49.830

You can use Inclusion-Exclusion for this. Here's how:

The letters comprise a multiset:

$$\mathfrak{W}_m=\{M^1,I^4,S^4,P^2\}$$

Firstly mark the letters with suffixes thus making them distinguishable (we will divide out duplicate permutations in the result). Thus we have the set:

$$\mathfrak{W}_s=\{M,I_1,I_2,I_3,I_4,S_1,S_2,S_3,S_4,P_1,P_2\}$$

We have, therefore, 4 letters of different kinds.

Then the inclusion-exclusion principle gives (after dividing out permutations of identical letters):

$$\#(\text{valid permutations of $\mathfrak{W}_m$})=\frac{1}{1!4!^22!}\sum_{k=0}^{7} (-1)^kS_k\tag{1}$$

where $S_k$ is the sum of all cardinalities of subset intersections of permutations of $\mathfrak{W}_s$ with at least $k$ ordered adjacent letter pairs of the same kind.

It should be clear that a letter kind with $n$ letters may have $r$ pairs of ordered adjacent letters selected in $\binom{n-1}{r}\frac{n!}{(n-r)!}$ ways: Out of $n!$ permutations we choose $r$ of the $n-1$ adjacent pairs in each. Then divide by the $(n-r)!$ arrangements of the $n-r$ blocks of letters which produce identical selections. Hence:

$$\begin{align}(-1)^kS_k=&(11-k)!\times\\[1ex] &\sum_{k=r_1+r_2+r_3} (-1)^{r_1}\binom{3}{r_1}\frac{4!}{(4-r_1)!}(-1)^{r_2}\binom{3}{r_2}\frac{4!}{(4-r_2)!}(-1)^{r_3}\binom{1}{r_3}\frac{2!}{(2-r_3)!}\, .\end{align}$$

We can see that $(-1)^kS_k$ is equivalent to the $x^{11-k}$ coefficient of the product

$$(11-k)!x\left(\binom{3}{0}\frac{4!}{4!}x^4-\binom{3}{1}\frac{4!}{3!}x^3+\binom{3}{2}\frac{4!}{2!}x^2-\binom{3}{3}\frac{4!}{1!}x\right)^2\left(\binom{1}{0}\frac{2!}{2!}x^2-\binom{1}{1}\frac{2!}{1!}x\right)\, .$$

Thus $(1)$ becomes (using the "find $x^m$ coefficient operator" $[x^m]$):

$$\begin{align}&\left(\sum_{k=0}^{7}(11-k)![x^{11-k}]\right)x\left(\frac{1}{24}x^4-\frac{1}{2}x^3+\frac{3}{2}x^2-x\right)^2\left(\frac{1}{2}x^2-x\right)\\[1ex] &=\frac{1}{1152} \, 11! - \frac{13}{576} \, 10! + \frac{11}{48} \, 9! - \frac{7}{6} \, 8! + \frac{77}{24} \, 7! - \frac{19}{4} \, 6! + \frac{7}{2} \, 5! - 4!\, ,\end{align}$$

which you may verify is $2016$.

Permutations of MISSISSIPPI not containing adjacent identical letters

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