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I have found something very interesting in my eyes.

The following expression $$\frac{\prod\limits_{i=n+1}^{2n-1}i}{\prod\limits^{n-1}_{i=2}i}$$

(I think) always gives a natural number. Now, of course, I am interested in why this is so and I wanted to ask if someone has an approach for a proof...

Thank you :)

(And yes, I know that I should explain my proof so you can verify it, but I have no approach for an proof, so if you tell me which proof-method I should use I would try it on my own of course :) )

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    $\begingroup$ This is an expression, not an equation. $\endgroup$ – barak manos Jan 2 '17 at 13:07
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    $\begingroup$ It is $^{2n-1}C_{n-1}$ $\endgroup$ – 8hantanu Jan 2 '17 at 13:09
  • $\begingroup$ @WiCK3DPOiSON: OK... So why is $\binom{2n-1}{n-1}$ an integer then? $\endgroup$ – barak manos Jan 2 '17 at 13:10
  • $\begingroup$ @Dr. Sonnhard Graubner I've made a mistake then I've deleted my hint ;) Thanks. $\endgroup$ – Olivier Oloa Jan 2 '17 at 13:18
  • $\begingroup$ Use: 1. $$\text{P}_1=\prod_{\text{m}=1+\text{n}}^{2\text{n}-1}\text{m}=\frac{\left(2\text{n}-1\right)!}{\text{n}!}$$ 2. $$\text{P}_2=\prod_{\text{m}=2}^{\text{n}-1}\text{m}=\left(\text{n}-1\right)!$$ So: $$\frac{\text{P}_1}{\text{P}_2}=\frac{\left(\frac{\left(2\text{n}-1\right)!}{\text{n}!}\right)}{\left(\text{n}-1\right)!}=\frac{1}{\text{n}!}\cdot\frac{\left(2\text{n}-1\right)!}{\left(\text{n}-1\right)!}=\frac{1}{2}\binom{2\text{n}}{\text{n}}$$ $\endgroup$ – Jan Jan 2 '17 at 13:32
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We can see that $$n!(\text {Numerator}) =1\times 2\times \cdots (2n-1) =(2n-1)! $$ Also , $$\text {Denominator} =(n-1)! $$ as $1!=1$. So, $$\frac {\text {Numerator}}{\text {Denominator}} =\frac {(2n-1)!}{n!(n-1)!} $$ With a little knowledge of binomial coefficients, we can see that this expression ( not an equation) simplifies to $$\binom {2n-1}{n-1} $$ Hope it helps.


Edit: For your question as to why this fraction is an integer, see this post.

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  • $\begingroup$ Okay, thanks! One last expression: In your last equation, did you forget the break, or is it a vector? $\endgroup$ – Gykonik Jan 2 '17 at 13:36
  • $\begingroup$ I am sorry but I can't understand which is the last equation. $\endgroup$ – Rohan Jan 2 '17 at 13:37
  • $\begingroup$ experssion, im sorry... :D This one (2n-1 | n-1) (| means new line..) This is a Vector, isn't it? $\endgroup$ – Gykonik Jan 2 '17 at 13:41
  • $\begingroup$ Oh no no, that's how we write the binomial coefficients. $\endgroup$ – Rohan Jan 2 '17 at 13:42
  • $\begingroup$ Oh, okay. I never heard of them, so I guess, that I have to look it up :P $\endgroup$ – Gykonik Jan 2 '17 at 13:43
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above question reduces into the statement product of any r consecutive integers is divisible by r!. The product of n consecutive integers is divisible by n factorial

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$$ \begin{align} \frac{\prod\limits_{k=n+1}^{2n-1}k}{\prod\limits_{k=2}^{n-1}k} &=\frac{\frac{(2n-1)!}{n!}}{(n-1)!}\\ &=\frac{(2n-1)!}{n!(n-1)!}\\[12pt] &=\binom{2n-1}{n} \end{align} $$ and Binomial Coefficients are integers as computed in Pascal's Triangle.

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