1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Permutation question (math) [ ]

  1. Apr 25, 2005 #1
    permutation question (math) [urgent]

    image here

    please solve this problem for me. the correct answer is c.

    i put 10!/(4!6!), then i know i am suppose to divide/subtract something, but i dont know what. (i have never done this kind of problem before.)
     
  2. jcsd
  3. Apr 25, 2005 #2

    JamesU

    User Avatar
    Gold Member

    hell im stumped...
     
  4. Apr 25, 2005 #3
    You can just add up the ways to get to each vertex. Here are the first few (period represents a vertex, number before each vertex is the number of ways to get to it):
    Code (Text):

    1. 1. 1. 1. 1.  .  .
    1. 2. 3. 4. 5.  .  .
    1. 3. 6.        .  .
    1. 4.10.  .  .  .  .
     .  .  .  .  .  .  .
     
    See what I'm doing and why?
     
  5. Apr 25, 2005 #4
    somebody plz help
     
  6. Apr 25, 2005 #5
    but how u gonna get the numbers for the dots in the park!

    i did try to solvei t using pascal triangle thing...but i got stuck on the park thing...
     
  7. Apr 25, 2005 #6
    There are no dots in the park. And it's not exactly Pascal's triangle because of the park. Here's a hint (some more vertexes filled in):
    Code (Text):

    1. 1. 1. 1. 1.  .  .
    1. 2. 3. 4. 5. 6.  .
    1. 3. 6.       6.  .
    1. 4.10.10.  .  .  .
     .  .  .  .  .  .  .
     
     
  8. Apr 25, 2005 #7
    Im guessing the park is considered as one big square.
     
  9. Apr 25, 2005 #8
    No, the park is considered a blank. The number of ways to get to any vertex is equal to the sum of the number of ways to get to any of the immediately preceding vertices.
     
  10. Apr 25, 2005 #9
    why u have a 10 after the 10??? how the get the second 10?
     
  11. Apr 25, 2005 #10
    For example take the first 10 you get, at vertex (4, 3) by row, column. At vertex (3, 3) you have a 6, and at vertex (4, 2) you have a 4. You can get to the 10 one of two ways: through (3, 3) by going south, and through (4, 2) by going east. There are 6 ways to get to (3, 3) so there are 6 ways to get to (3, 3) and then go south. There are 4 ways to get to (4, 2) so there are 4 ways to get to (4, 2) and then go east. So you have 4 ways + 6 ways = 10 ways for vertex (4, 3).

    If a vertex only has one other vertex leading into it--say the other vertex has 7 ways to get to it--then how many ways can you get to that second vertex?
     
  12. Apr 25, 2005 #11
    oh....i think i got it....thanks for this "If a vertex only has one other vertex leading into it--say the other vertex has 7 ways to get to it--then how many ways can you get to that second vertex?"

    again..thanks alot!!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Permutation question (math) [ ]
  1. Math question (Replies: 2)

  2. Math questions (Replies: 1)

Loading...