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Basic combinations problem

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data

    If k people are seated in a random manner in a row containing n seats (n>k), what is the probability that the people will occupy k adjacent seats in the row?

    I realize that there are n choose k sets of k seats to be occupied, and that there are n-k+1 sets of k adjacent seats. So the probability that I'm looking for is:

    (n-k+1)/(n choose k)

    What I don't understand is how the above probability simplifies to:

    [(n-k+1)!k!]/n!

    Can someone please explain? Thanks.


    2. Relevant equations

    (nk) = n choose k = n!/[(n-k)!k!]


    3. The attempt at a solution

    (n-k+1)/(n choose k) = [(n-k+1)(n-k)!k!]/n!

    Not sure what to do from here.
     
  2. jcsd
  3. Apr 27, 2010 #2
    Look at (n-k+1)(n-k)!, can you simplify this product any?
     
    Last edited: Apr 27, 2010
  4. Apr 27, 2010 #3
    Other than finding a quotient, I've never had to manipulate/simplify factorials. Thinking about this again, since the number (n-k+1) is 1 greater than the number (n-k), (n-k)! times (n-k+1) must equal (n-k+1)!.
     
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