Derangement of selective letters in a string.

combinatorial-proofscombinatoricsderangementspermutations

I got to know about the derangemnt formula which can be found here:
These are links to some excellent answers to this very topic.

  1. Link
  2. A link to my previous question where some links related to this topic were also given.

So my problem is how to solve question when only some specific letters are asked to be repeated.

Like, consider the problems:

1)How many permutations of 1,… 8 are there in which no even number appears in its natural position?

( Yes I know this could be solved by principle of inclusion and exclusion but how to use the Rooks formula here?)

An excellent answer to this problem can be found on quora here.
(Using princeple of inclusion exclusion)

Or also one more Interesting problem

2)Find the derangements of "ABHIBHAV"

Yes, these question are solved by Principle of inclusion and exclusion but those require a quite a bit tougher level of understanding to decipher problem rightly.. and recently I got to know about Rooks theorem, by which I tried many problems, which got solved but they were all based on "all letter have to be deranged types" so here in these kind of problem the case is different so my question is basically how to use that formula in these cases.

Any refferences for further read/ extra typical problems links would be also highly appreciated.

Best Answer

For the question about permuting the the numbers $1$ through $8$, where the even numbers don't occupy their original positions, the chessboard is an $8\times8$ square with positions $(2,2), (4,4), (6,6), (8,8)$ missing, or blacked out. We want to compute the rook polynomial of the black chessboard.

One handy fact is that when we have two chessboards with no row or column in common, the the rook polynomial of their union is the product of theirs rook polynomials. In this case we have four one-cell black chessboards, no two of which have a cell in the same row or column. Each has rook polynomial $(1+x)$, so the black chessboard has rook polynomial $$(1+x)^4:=\sum_{k=0}^4a_kx^k$$

Now the number of admissible permutations is $$8!-a_17!+a_26!-a_35!+a_44!$$ since $a_k=0$ for $k>4$.

There is a really good chapter on rook polynomials in "Introduction to Combinatorial Mathematics" by C.L Liu. This book is long out of print, but you may be able to find it in a library, or pick it up used

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