Calculating the Value of Combinations with Induction

In summary, the conversation discusses the problem of finding the value of the series \sum (n k) from k=0 to n, where n and k are positive integers. The solution involves using the Binomial Theorem and proving it through induction. The final result is determined to be 2^n.
  • #1
SMA_01
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.
 
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  • #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
 
  • #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...?
 
  • #4
SMA_01 said:
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. :)
 

1. What is the value of combinations?

The value of combinations is determined by the number of possible unique arrangements that can be made from a given set of items or elements. It is often used in mathematics and statistics to calculate probabilities and outcomes.

2. How is the value of combinations calculated?

The value of combinations is calculated using the formula nCr = n! / r!(n-r)!, where n is the total number of items and r is the number of items in each combination. This formula can be used for both ordered and unordered combinations.

3. What is the difference between combinations and permutations?

Combinations and permutations are both ways of arranging items, but the main difference is that combinations do not take into account the order of the items, while permutations do. In combinations, the same items in a different order are considered the same combination, while in permutations they are considered different arrangements.

4. How can combinations be used in real-life situations?

Combinations are used in a variety of real-life situations, such as in lottery games to calculate the chances of winning, in password generation to create unique and secure combinations, and in genetics to determine the possible combinations of genes in offspring.

5. Can the value of combinations be negative?

No, the value of combinations cannot be negative. It is always a positive integer, as it represents the number of possible arrangements or outcomes. If the calculated value is negative, it means that the formula was not applied correctly.

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