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[tex]\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}=2^{n}[/tex]

Any help is appreciated.

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- Thread starter amcavoy
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- #1

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[tex]\sum_{k=0}^{n}\frac{n!}{k!\left(n-k\right)!}=2^{n}[/tex]

Any help is appreciated.

- #2

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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.....

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.....

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Thanks a lot for your help.

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matt grime

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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

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alexmcavoy@gmail.com said:

Thanks a lot for your help.

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 [tex]sin^2(x) + cos^2(x) = 1[/tex] 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,

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- #6

honestrosewater

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"Let S be a finite set with N elements. Then the powerset of S (that is the set of all subsets of S ) contains 2^N elements. ..."

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matt grime

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

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

- #10

matt grime

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

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

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

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Gokul43201

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...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).matt grime said:combinatorially [itex]\binom{n}{k}[/itex] 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....

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

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