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zepdrix
 one year ago
(Combinatorics)
In how many ways can
five indistinguishable rooks
be placed on an 8by8 chess board
so that no rook can attack another and
neither the first row nor the first column is empty?
zepdrix
 one year ago
(Combinatorics) In how many ways can five indistinguishable rooks be placed on an 8by8 chess board so that no rook can attack another and neither the first row nor the first column is empty?

This Question is Closed

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3The answer key says \(\rm =\left(\begin{matrix}7\\4\end{matrix}\right)^2 4!+7!\left(\begin{matrix}6\\3\end{matrix}\right)^2 3!\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3I'm seeing addition in the answer, so I assume I'm probably need to break this into cases.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3It's just not quite working out for me though, hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Place one rook into the first row / first column. Then place the other 4 rooks into 7 rows \(\left(\begin{matrix}7\\4\end{matrix}\right)\) Then these 4 rooks have 7 columns to choose from \(\large\rm P(7,4)\) And I did some calculations and it turns out that:\[\large\rm \left(\begin{matrix}7\\4\end{matrix}\right)^2 4!=\left(\begin{matrix}7\\4\end{matrix}\right)P(7,4)\]Oo ok good! I'm on the right track here.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3I similarly messed with the answer key to the second part and turned it into:\[\large\rm 7!\left(\begin{matrix}6\\3\end{matrix}\right)P(6,3)\]Hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Ooo Google to the rescue! yay, found something

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3http://jeremycue.com/wpcontent/uploads/2013/06/10MoreCountingProblems.pdf Problem 7

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Ok I guess I setup the first CASE correctly, placing the rook in the upper left corner. CASE 2 should be no rook in upper left? Hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Case 2: So when we look at the combinations of rows, we'll have to reserve one rook for the top row. So we only have 4 other rooks to work with. Hmm it looks like 3 in the answer though.. hmmm thinking

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Oh oh so if a rook is not in the upper left corner, it means i need to reserve a rook for the first row, and another for the first column separately... ok ok ok

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3The rook in the first row has 7 options. The rook in the first column has 7 options. Leaving me with 3 rooks to place in 6 rows \(\left(\begin{matrix}6\\3\end{matrix}\right)\) and permutate them in the columns \(\large\rm P(6,3)\) Oooo yay I think I got it! \(\large\rm case2=7\cdot7\left(\begin{matrix}6\\3\end{matrix}\right)P(6,3)\) Which matches the solution from that website, but not the book. I think it's more likely the book is wrong. Man I deserve a medal for that workout _

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1I agree with your logic: Case 1: occupy square A1 (lower left corner): 4 rooks left, with 49 squares, 36 squares, 25 squares, 16 squares, etc. N1=49*36*25*16/4!=29400 ........=C(7,4)^2*P(7,4) Case 2: occupy one of A2A8, AND B1H1, 3 rooks left: N1=7*7*36*25*16/3!=117600............=7^2*C(6,3)^2*P(6,3) Total N1+N2=147000 ways This matches the given answer if the given answer were: \(\rm =\left(\begin{matrix}7\\4\end{matrix}\right)^2 4!+\color{red}{7^2 }\left(\begin{matrix}6\\3\end{matrix}\right)^2 3!\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.3Doh!! It is a 7^2 in the book :P Man I'm a doofus!
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