• Support PF! Buy your school textbooks, materials and every day products Here!

Value of combinations?

  • Thread starter SMA_01
  • Start date
  • #1
218
0

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
3,812
92
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
218
0
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
3,812
92
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. :)
 

Related Threads for: Value of combinations?

Replies
10
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
569
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
1
Views
368
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
2
Views
2K
Top