Combinatorics Class - Sum Question

theRukus
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Homework Statement


For any positive integer n determine:

\sum\limits^n_{i=0} \frac{1}{i!(n-i)!}

Homework Equations



I don't really know where to start.. Up until this point we've just been doing permutations, combinations, and determining the coefficient of a certain term in the expansion of a polynomial. There aren't any examples like this question in the text, and so I am unsure as to what sort of an answer they are looking for... Are they just looking for a general formula (not a sum) for the answer, with n as a variable? Cheers for any direction!

The Attempt at a Solution

 
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Hint: does this look familiar?

\frac{n!}{i!(n-i)!}
 
So the answer I'm looking for is

\frac{\dbinom{n}{i}}{n!}

Correct?
 
Or will it be

\sum\limits^n_{i=0} \dfrac{\dbinom{n}{i}}{n!}

I'm confused as to whether the sum is still involved.
 
you should find the following sum:

\frac{1}{n!}*\sum \frac{n!}{i! (n-1)!}
 
theRukus said:
Or will it be

\sum\limits^n_{i=0} \dfrac{\dbinom{n}{i}}{n!}

I'm confused as to whether the sum is still involved.

Of course the sum is still involved. The final answer must be in terms of n alone: it cannot contain "i", since all values of i have been summed over. Anyway, just multiplying and dividing by n! does not magically get rid of the sum.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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