Why does $binom2nn = prod_i=1^n (fracn+ii )$
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I cannot understand why are you able to generalize this like this
$$binom2nn = frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1 =prod_i=1^n (fracn+ii) $$
I get that $$binom2nn = frac2n!n! (n-n)! $$ and therefore this is correct $$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1 $$
However, how do you get to that product generalization?
What would you suggest I read to understand it?
binomial-coefficients
add a comment |Â
up vote
-1
down vote
favorite
I cannot understand why are you able to generalize this like this
$$binom2nn = frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1 =prod_i=1^n (fracn+ii) $$
I get that $$binom2nn = frac2n!n! (n-n)! $$ and therefore this is correct $$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1 $$
However, how do you get to that product generalization?
What would you suggest I read to understand it?
binomial-coefficients
1
It’s only a way to rewrite the expression
– gimusi
Jul 30 at 5:32
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I cannot understand why are you able to generalize this like this
$$binom2nn = frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1 =prod_i=1^n (fracn+ii) $$
I get that $$binom2nn = frac2n!n! (n-n)! $$ and therefore this is correct $$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1 $$
However, how do you get to that product generalization?
What would you suggest I read to understand it?
binomial-coefficients
I cannot understand why are you able to generalize this like this
$$binom2nn = frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1 =prod_i=1^n (fracn+ii) $$
I get that $$binom2nn = frac2n!n! (n-n)! $$ and therefore this is correct $$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1 $$
However, how do you get to that product generalization?
What would you suggest I read to understand it?
binomial-coefficients
asked Jul 30 at 5:09
Luis Esparza LeedMx
81
81
1
It’s only a way to rewrite the expression
– gimusi
Jul 30 at 5:32
add a comment |Â
1
It’s only a way to rewrite the expression
– gimusi
Jul 30 at 5:32
1
1
It’s only a way to rewrite the expression
– gimusi
Jul 30 at 5:32
It’s only a way to rewrite the expression
– gimusi
Jul 30 at 5:32
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
accepted
We simply have
$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$
Ohhh I see, I feel so embarrassed, is so obvious now. Thank you very much.
– Luis Esparza LeedMx
Jul 30 at 7:06
You are welcome! Don’t feel embarrassed, any doubt deserves a clarification. Bye
– gimusi
Jul 30 at 7:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We simply have
$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$
Ohhh I see, I feel so embarrassed, is so obvious now. Thank you very much.
– Luis Esparza LeedMx
Jul 30 at 7:06
You are welcome! Don’t feel embarrassed, any doubt deserves a clarification. Bye
– gimusi
Jul 30 at 7:35
add a comment |Â
up vote
1
down vote
accepted
We simply have
$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$
Ohhh I see, I feel so embarrassed, is so obvious now. Thank you very much.
– Luis Esparza LeedMx
Jul 30 at 7:06
You are welcome! Don’t feel embarrassed, any doubt deserves a clarification. Bye
– gimusi
Jul 30 at 7:35
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We simply have
$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$
We simply have
$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$
answered Jul 30 at 5:14
gimusi
64.5k73482
64.5k73482
Ohhh I see, I feel so embarrassed, is so obvious now. Thank you very much.
– Luis Esparza LeedMx
Jul 30 at 7:06
You are welcome! Don’t feel embarrassed, any doubt deserves a clarification. Bye
– gimusi
Jul 30 at 7:35
add a comment |Â
Ohhh I see, I feel so embarrassed, is so obvious now. Thank you very much.
– Luis Esparza LeedMx
Jul 30 at 7:06
You are welcome! Don’t feel embarrassed, any doubt deserves a clarification. Bye
– gimusi
Jul 30 at 7:35
Ohhh I see, I feel so embarrassed, is so obvious now. Thank you very much.
– Luis Esparza LeedMx
Jul 30 at 7:06
Ohhh I see, I feel so embarrassed, is so obvious now. Thank you very much.
– Luis Esparza LeedMx
Jul 30 at 7:06
You are welcome! Don’t feel embarrassed, any doubt deserves a clarification. Bye
– gimusi
Jul 30 at 7:35
You are welcome! Don’t feel embarrassed, any doubt deserves a clarification. Bye
– gimusi
Jul 30 at 7:35
add a comment |Â
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1
It’s only a way to rewrite the expression
– gimusi
Jul 30 at 5:32