A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

zepdrix

  • one year ago

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

  • This Question is Closed
  1. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    The answer key says \(\rm =\left(\begin{matrix}7\\4\end{matrix}\right)^2 4!+7!\left(\begin{matrix}6\\3\end{matrix}\right)^2 3!\)

  2. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I'm seeing addition in the answer, so I assume I'm probably need to break this into cases.

  3. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    It's just not quite working out for me though, hmm

  4. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Place 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.

  5. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    I 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

  6. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Ooo Google to the rescue! yay, found something

  7. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    http://jeremycue.com/wp-content/uploads/2013/06/10-More-Counting-Problems.pdf Problem 7

  8. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Ok 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

  9. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Case 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

  10. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Oh 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

  11. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    The 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 -_-

  12. mathmate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I 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 A2-A8, AND B1-H1, 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: \(\rm =\left(\begin{matrix}7\\4\end{matrix}\right)^2 4!+\color{red}{7^2 }\left(\begin{matrix}6\\3\end{matrix}\right)^2 3!\)

  13. zepdrix
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Doh!! It is a 7^2 in the book :P Man I'm a doofus!

  14. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.