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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.

Piyush Sawarkar
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  • Do you know how to compute the rook polynomial in the first case? If so, what is the difficulty? – saulspatz Jun 23 '20 at 17:07
  • @saulspatz Yes actually I know how to form the polynomial and then we multiply but the case is, here all are not to be deranged then say if I have "abbc" then by rooks formula to derange all letters it's simply L2^1*L1^2 and then further is just multiplying polynomials. But now what if I have a question where only some are free say in the same hypothetical question "abbc" say if only "ab" is supposed to be deranged then how to used in these cases...I am sure I am missing some thing but still an intuitional help would also be helpful for me to grasp the concept and apply in these cases too.. – Piyush Sawarkar Jun 23 '20 at 17:12
  • That comment isn't clear to me. In case a) the rook polynomial is $(1+x)^4$. Do you know how to get that? – saulspatz Jun 23 '20 at 17:15
  • $(1+x)^4$ ? Can you plz further explain how is it coming? okay...then it seems I don't know how to apply bcoz as per me what I know is consider simple problem of deranging "abc" ..as per me the answer is L1L1L1 where L1=(x-1) then final answer comes by expanding (x-1)^3 =x^3-3x^2+3x-1; Now to evaluate this.. answee would be = 3!-3(2!)+3(1!)-1 Is this right way how am I supposed to do ..bcoz I have solved problems this way only (yes I don't know real intuition behind these formula) – Piyush Sawarkar Jun 23 '20 at 17:21
  • Please use MathJax to format your posts, and break what you are trying to say up into sentences. I don't understand what you are asking me. Instead of trying to put this in comments, please edit your question to clarify what you are asking. What does "abc" have to do with permuting the numbers from $1$ to $8$? – saulspatz Jun 23 '20 at 17:24
  • Okay sorry for inconvenience but can you explain how you got $(1+x)^4$ ? And that "abc" example was just an example of how we derange those 3 elements, it has nothing to do with main question.. – Piyush Sawarkar Jun 23 '20 at 17:29

1 Answers1

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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

saulspatz
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  • thanks for the reference! – karakfa Jun 23 '20 at 20:03
  • Can you please elaborate a bit more about how is this (1+x) term coming I am not understanding ,sorry if it is a very basic question. Thank you for answering ! – Piyush Sawarkar Jun 23 '20 at 22:57
  • @PiyushSawarkar The rook polynomial of a chessboard is $\sum a_kx^k$ where $a_k$ is the number of ways to place $k$ non-attacking rooks on the chessboard. If the chessboard consists of a single square, then $a_0=a_1=1$. – saulspatz Jun 23 '20 at 23:09
  • Oh yes!! thank you Sir @saulspatz now I understood the Crux of it.. Also can you tell me how should I consider the chess board arrangement for second problem which I have asked ..like how to proceed in those cases.. a help would be highly appreciated, thank you! (I have edited the second question and changed it) – Piyush Sawarkar Jun 24 '20 at 00:06
  • Thank you sir I finally got all the concepts @saulspatz – Piyush Sawarkar Jun 24 '20 at 14:55
  • As to the second part, I'm not sure I understand what you mean by "derangements. If you want the permutations of "ABHIbhaV" such that neither b nor B can occupy the second or fifth place, neither H nor h can occupy the third or sixth, neither A nor a can occupy the first or seventh, I cannot occupy the fourth, and V cannot occupy the eighth, then you just need to blacken the corresponding squares on an $8\times8$ chessboard, compute the rook polynomial, and apply the formula. If you permute the rows and columns of the chessboard, the rook polynomial is unchanged ... – saulspatz Jun 24 '20 at 15:07
  • ... so we can make $3$ $2\times2$ and $2$ $1\times1$ black squares, with no row of column in common, giving a rook polynomial of $(1+4x+2x^2)^3(1+x)^2$. – saulspatz Jun 24 '20 at 15:09