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Value of combinations?

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Let n, k be in the set of all positive integers including zero. Define (n k)=n!/k!(n-k)!
    Determine the value of [itex]\sum[/itex] (n k) from k=0 to n. Determine the value of [itex]\sum[/itex] (n k) from k=0 to n.



    3. The attempt at a solution

    I tried evaluating for some n value:

    (n 0)= 1

    (n 1)= n

    (n 2)= (1/2!)n(n-1)

    (n 3)= (1/3!)n(n-1)(n-2)


    So, generally= 1+n+......+[(1/k!)(n)*.....*(n-(n-k))]+1

    Am I on the right track? How can I determine the value?


    Any help is appreciated.
     
  2. jcsd
  3. Sep 15, 2012 #2
    I don't think that this problem requires use of Calculus, it can be simply done using the Binomial Theorem.

    What is (x+a)n? Write that and try to find the values of x and a by comparing the formula with the series you got i.e. with 1+n+......+[(1/k!)(n)*.....*(n-(n-k))].

    http://en.wikipedia.org/wiki/Binomial_theorem
     
  4. Sep 15, 2012 #3
    Pranav-Arora: Thanks. I'm going to have to prove it using induction. Is there anyway other than the binomial theorem to reach the result? I know it will equal 2^n, but I was wondering if you can compute without using the binomial theorem...?
     
  5. Sep 15, 2012 #4
    I am not sure if i know of any other method, you should better wait for others members to look at this thread. :)
     
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