Inequality in binomial coefficient [closed]












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$begingroup$


How to prove
$$binom{2n+x}{n} binom{2n-x}{n} leqbinom{2n}{n}^2$$



I tried by expanding. But then not able to proceed.






binomial-coefficients







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closed as off-topic by Alexander Gruber♦ Jan 30 at 6:03


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


    How to prove
    $$binom{2n+x}{n} binom{2n-x}{n} leqbinom{2n}{n}^2$$



    I tried by expanding. But then not able to proceed.






    binomial-coefficients







    share|cite|improve this question









    $endgroup$



    closed as off-topic by Alexander Gruber♦ Jan 30 at 6:03


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1





      $begingroup$


      How to prove
      $$binom{2n+x}{n} binom{2n-x}{n} leqbinom{2n}{n}^2$$



      I tried by expanding. But then not able to proceed.






      binomial-coefficients







      share|cite|improve this question









      $endgroup$




      How to prove
      $$binom{2n+x}{n} binom{2n-x}{n} leqbinom{2n}{n}^2$$



      I tried by expanding. But then not able to proceed.






      binomial-coefficients




      binomial-coefficients






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 30 at 4:40









      mavericmaveric

      89412




      89412




      closed as off-topic by Alexander Gruber♦ Jan 30 at 6:03


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Alexander Gruber♦ Jan 30 at 6:03


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          $begingroup$

          I think it means $0leq xleq n$.



          We need to prove that
          $$tfrac{(2n)^2(2n-1)^2...(n+1)^2}{n!^2}geqtfrac{(2n+x)(2n+x-1)...(n+x+1)}{n!}cdottfrac{(2n-x)(2n-x-1)...(n-x+1)}{n!}$$ or
          $$(2n)^2(2n-1)^2...(n+1)^2geq((2n)^2-x^2)((2n-1)^2-x^2)...((n+1)^2-x^2),$$ which is obvious.






          share|cite|improve this answer









          $endgroup$




















            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            I think it means $0leq xleq n$.



            We need to prove that
            $$tfrac{(2n)^2(2n-1)^2...(n+1)^2}{n!^2}geqtfrac{(2n+x)(2n+x-1)...(n+x+1)}{n!}cdottfrac{(2n-x)(2n-x-1)...(n-x+1)}{n!}$$ or
            $$(2n)^2(2n-1)^2...(n+1)^2geq((2n)^2-x^2)((2n-1)^2-x^2)...((n+1)^2-x^2),$$ which is obvious.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              I think it means $0leq xleq n$.



              We need to prove that
              $$tfrac{(2n)^2(2n-1)^2...(n+1)^2}{n!^2}geqtfrac{(2n+x)(2n+x-1)...(n+x+1)}{n!}cdottfrac{(2n-x)(2n-x-1)...(n-x+1)}{n!}$$ or
              $$(2n)^2(2n-1)^2...(n+1)^2geq((2n)^2-x^2)((2n-1)^2-x^2)...((n+1)^2-x^2),$$ which is obvious.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                I think it means $0leq xleq n$.



                We need to prove that
                $$tfrac{(2n)^2(2n-1)^2...(n+1)^2}{n!^2}geqtfrac{(2n+x)(2n+x-1)...(n+x+1)}{n!}cdottfrac{(2n-x)(2n-x-1)...(n-x+1)}{n!}$$ or
                $$(2n)^2(2n-1)^2...(n+1)^2geq((2n)^2-x^2)((2n-1)^2-x^2)...((n+1)^2-x^2),$$ which is obvious.






                share|cite|improve this answer









                $endgroup$



                I think it means $0leq xleq n$.



                We need to prove that
                $$tfrac{(2n)^2(2n-1)^2...(n+1)^2}{n!^2}geqtfrac{(2n+x)(2n+x-1)...(n+x+1)}{n!}cdottfrac{(2n-x)(2n-x-1)...(n-x+1)}{n!}$$ or
                $$(2n)^2(2n-1)^2...(n+1)^2geq((2n)^2-x^2)((2n-1)^2-x^2)...((n+1)^2-x^2),$$ which is obvious.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 5:17









                Michael RozenbergMichael Rozenberg

                109k1896201




                109k1896201















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