# Sum of Combinations

1. Jul 26, 2005

### amcavoy

I was just wondering how you would prove the following:

$$\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}=2^{n}$$

Any help is appreciated.

2. Jul 27, 2005

### Townsend

Its easy to see once you know where to look. Look at pascals triangle. What is the value of the sum of all the numbers in each row equal to? Perhaps 2^n....where n is the row...

Now what does that binomial theorem have to say about combinations and pascals triangle again?

Do you see where to go from here?

Just so you know, what I am talking about is how you can show that your equation is believable but to prove it you would have to show all the details.....

Last edited: Jul 27, 2005
3. Jul 27, 2005

### amcavoy

I see what you are saying about Pascal's Triangle. However, I would like to know if there is a way to prove this symbolically, rather than just seeing that it works. Does anyone have any suggestions?

Thanks a lot for your help.

4. Jul 27, 2005

### matt grime

combinatorially $\binom{n}{k}$ which is the summand is the number of ways of choosing k objects (order unimportant) from n or the number of subsets fo size k of n objects.

you are adding these up from 0 to n.

so you are finding the total number of subsets of a set of size n. this is obvisouly 2^n

it may also be proved by induction or by noting it is the binomial expansion of (x+y)^n for x=y=1

5. Jul 27, 2005

### Townsend

If what you're asking is for a way to prove the identity directly.....good luck.

The first time I realized that sum was equal to 2^n I went to trying to prove the identity like what you're asking. I never could....

But good luck to you.

As a side note, proving it the way matt grime said is just as good as any other way. After all, can you prove $$sin^2(x) + cos^2(x) = 1$$ symbolically? You could...perhaps....but it's no better than a geometric argument. And I think the geometric proof is more useful in the sense that you have a geometric understanding of why the identity is true.

Regards,

Last edited: Jul 27, 2005
6. Jul 27, 2005

### honestrosewater

7. Jul 27, 2005

### matt grime

hopefully i gave a solution that is more explanatory than "jsut look at pascal's triangle" ie it explains why the rows add up to 2^n rather than just saying they do.

8. Jul 27, 2005

### amcavoy

Yes of course. Thanks a lot for your help everyone.

9. Jul 29, 2005

### Pietjuh

You can also see that the sum is just equal to $$(1+x)^n$$ evaluated at x=1

10. Jul 31, 2005

### matt grime

but i said that one too in one of the three proofs i gave...

11. Jul 31, 2005

### Townsend

Yes, but to the casual observer such things may not be so clear....thus, it never hurts to be explicit from time to time.

12. Jul 31, 2005

### Gokul43201

Staff Emeritus
...to find the number of ways of selecting any number of objects out of n given objects. (ie: you can pick 0 objects or 1 object or 2 objects or...or all n objects).

Alternatively, you can think of this as assigning to each of the n objects, one of 2 labels, namely "chosen" and "not chosen". The total number of ways of assigning labels is hence 2^n, and this is exactly the process of choosing any number of objects.

Last edited: Jul 31, 2005
13. Jul 31, 2005

### LittleWolf

You can prove Sum(nCk,k=0,1..n)=2^n as a straight summation problem by induction on n. It works for n=0 then assume it is true for n-1. Since 2^(n-1)+2^(n-1) = 2^n expand out the summations and rearrange terms and show (n-1)C(k-1)+(n-1)Ck=nCk Where nCk=n!/k!/(n-k)! and note that nC0=(n-1)C0=nCn=(n-1)C(n-1)=1.