Why does $\binom{2n}{n} = \prod_{i=1}^n (\frac{n+i}{i} )$

Question

I cannot understand why are you able to generalize this like this

$$\binom{2n}{n} = \frac{2n × (2n−1) × ... × (n+ 2) × (n+ 1)}{n ×(n−1)×...×2×1} =\prod_{i=1}^n (\frac{n+i}{i}) $$

I get that $$\binom{2n}{n} = \frac{2n!}{n! (n-n)!} $$ and therefore this is correct $$\frac{2n × (2n−1) × ... × (n+ 2) × (n+ 1)}{n ×(n−1)×...×2×1} $$

However, how do you get to that product generalization?

What would you suggest I read to understand it?

It’s only a way to rewrite the expression – user Jul 30 '18 at 05:32 — user, Jul 30 '18 at 05:32

score 2 · Accepted Answer · answered Jul 30 '18 at 05:14

2

We simply have

$$\frac{2n × (2n−1) × ... × (n+ 2) × (n+ 1)}{n ×(n−1)×...×2×1}=\frac{2n}{n} × \frac{2n−1}{n-1} × \frac{2n−2}{n-2}× \ldots × \frac{n+1}{1}= \prod_{i=1}^n \left(\frac{n+i}{i}\right)$$

answered Jul 30 '18 at 05:14

user

154,566

Ohhh I see, I feel so embarrassed, is so obvious now. Thank you very much. – Luis Esparza LeedMx Jul 30 '18 at 07:06
You are welcome! Don’t feel embarrassed, any doubt deserves a clarification. Bye – user Jul 30 '18 at 07:35

Why does $\binom{2n}{n} = \prod_{i=1}^n (\frac{n+i}{i} )$

1 Answers1