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




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  • 1




    It’s only a way to rewrite the expression
    – gimusi
    Jul 30 at 5:32














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




share|cite|improve this question















  • 1




    It’s only a way to rewrite the expression
    – gimusi
    Jul 30 at 5:32












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




share|cite|improve this question











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





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









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












  • 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










1 Answer
1






active

oldest

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up vote
1
down vote



accepted










We simply have



$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$






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  • 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










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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



$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$






share|cite|improve this answer





















  • 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














up vote
1
down vote



accepted










We simply have



$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$






share|cite|improve this answer





















  • 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












up vote
1
down vote



accepted







up vote
1
down vote



accepted






We simply have



$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$






share|cite|improve this answer













We simply have



$$frac2n × (2n−1) × ... × (n+ 2) × (n+ 1)n
×(n−1)×...×2×1=frac2nn × frac2n−1n-1 × frac2n−2n-2× ldots × fracn+11= prod_i=1^n left(fracn+iiright)$$







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











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
















  • 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












 

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