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I have $\sum_{k=0}^{p-1}\frac{(p-1)!}{(p-1-k)!k!}$. Whereby $p$ is prime.

how can I say that the fraction's result will be an integer?

I tried it with numbers and yes it works. So I tried two approaches:

My first try was to write the factorial as a product,which means, $\frac{\prod_{k=1}^{p-1}k}{\prod_{i=1}^{p-1-k}i\prod_{j=1}^{k}j}$ but I didn't get much further and the second one was to try and use some properties of the factorial and I wrote it as: $\frac{1!(p-2)!(p-1)}{k!(p-1-k)(p-2-k)!}$.

In both cases I tried to expand them and stuff but didn't come to any conclusion. If I expand them I do understand why it must work, but I can't put it in some mathematical decent way.

I would like to solve it alone but I would be thankful for some tip, or maybe for some other property that I am missing, so that then I can continue it.

Thanks in advance

Annalisa

Annalisa
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1 Answers1

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This is the binomial expansion $$(a+b)^n = \sum\limits_{k=0}^n{n \choose k}a^kb^{n-k}$$ with $n=p-1$ and $a=b=1$. (Recall the symbol ${n\choose k} = \frac{n!}{k!(n-k)!}$ is an integer when $n$ and $k$ are, as it is the number of combinations of $n$ objects taken $k$ at a time.)

So your sum is $2^{p-1}$.

MPW
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  • I had absolutely no idea that it was known that the binomial coefficient was an integer. It helped a lot. Thanks. I also wanted to ask you something, let's say $a$ is not 1, but another integer, am I allowed to take it out of the sum? Because it is to the power of k and $k$ is changing. – Annalisa Feb 08 '21 at 12:51
  • It follows from the fact that the product of $k$ consecutive positive integers is divisible by $k!$ (prove that!). – MPW Feb 08 '21 at 12:56
  • You can’t bring anything with a “$k$” in it out of the sum – MPW Feb 08 '21 at 13:15