Value of combinations?

  • Thread starter SMA_01
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  • #1
SMA_01
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Homework Statement



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.



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.
 

Answers and Replies

  • #2
Saitama
4,244
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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
 
  • #3
SMA_01
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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...?
 
  • #4
Saitama
4,244
93
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...?

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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