Is it possible to show












0














As part of a larger proof I need the following to be true in order to make it work,



$(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$



Can anyone give me a hand?






binomial-coefficients







share|cite|improve this question



























    0














    As part of a larger proof I need the following to be true in order to make it work,



    $(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$



    Can anyone give me a hand?






    binomial-coefficients







    share|cite|improve this question

























      0












      0








      0







      As part of a larger proof I need the following to be true in order to make it work,



      $(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$



      Can anyone give me a hand?






      binomial-coefficients







      share|cite|improve this question













      As part of a larger proof I need the following to be true in order to make it work,



      $(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$



      Can anyone give me a hand?






      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 Nov 28 at 17:13









      I suck at Maths

      178




      178






















          2 Answers
          2






          active

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          1














          It’s just really a sequence of simplifications.



          $$(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$$



          $$color{blue}{(a+1)}frac{b!}{color{blue}{(a+1)!}(b-(a+1))!}x^{a+1}(1-x)^{b-a-1}=color{green}{a}frac{b!}{color{green}{a!}(b-a)!}x^a(1-x)^{b-a}$$



          $$frac{b!}{a!(b-a-1)!}color{blue}{x^{a+1}}color{green}{(1-x)^{b-a-1}}=frac{b!}{(a-1)!(b-a)!}color{blue}{x^a}color{green}{(1-x)^{b-a}}$$



          $$frac{color{blue}{b!}}{color{green}{a!}color{purple}{(b-a-1)!}}x=frac{color{blue}{b!}}{color{green}{(a-1)!}color{purple}{(b-a)!}}(1-x)$$



          $$frac{x}{a}=frac{1-x}{b-a} implies x(b-a)=a(1-x) implies bx-ax=a-ax implies bx = a$$






          share|cite|improve this answer





























            2














            Unless $x=0$ or $x=1$, $$frac{(a+1)b!}{(a+1)!(b-a-1)!}x^{a+1}(1-x)^{b-a-1}=frac{ab!}{a!(b-a)!}x^a(1-x)^{b-a}$$



            simplifies to



            $$(b-a)x=a(1-x).$$






            share|cite|improve this answer





















            • It also further simplifies to bx = a.
              – Ofya
              Nov 28 at 17:25










            • Since we know b > a, we can say that 1>x>0 as well
              – Ofya
              Nov 28 at 17:25










            • Actually, in the context of the problem earlier I showed that $a=bx$, so this solves my problem, thank you!
              – I suck at Maths
              Nov 28 at 17:36











            Your Answer





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            2 Answers
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            2 Answers
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            active

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            active

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            1














            It’s just really a sequence of simplifications.



            $$(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$$



            $$color{blue}{(a+1)}frac{b!}{color{blue}{(a+1)!}(b-(a+1))!}x^{a+1}(1-x)^{b-a-1}=color{green}{a}frac{b!}{color{green}{a!}(b-a)!}x^a(1-x)^{b-a}$$



            $$frac{b!}{a!(b-a-1)!}color{blue}{x^{a+1}}color{green}{(1-x)^{b-a-1}}=frac{b!}{(a-1)!(b-a)!}color{blue}{x^a}color{green}{(1-x)^{b-a}}$$



            $$frac{color{blue}{b!}}{color{green}{a!}color{purple}{(b-a-1)!}}x=frac{color{blue}{b!}}{color{green}{(a-1)!}color{purple}{(b-a)!}}(1-x)$$



            $$frac{x}{a}=frac{1-x}{b-a} implies x(b-a)=a(1-x) implies bx-ax=a-ax implies bx = a$$






            share|cite|improve this answer


























              1














              It’s just really a sequence of simplifications.



              $$(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$$



              $$color{blue}{(a+1)}frac{b!}{color{blue}{(a+1)!}(b-(a+1))!}x^{a+1}(1-x)^{b-a-1}=color{green}{a}frac{b!}{color{green}{a!}(b-a)!}x^a(1-x)^{b-a}$$



              $$frac{b!}{a!(b-a-1)!}color{blue}{x^{a+1}}color{green}{(1-x)^{b-a-1}}=frac{b!}{(a-1)!(b-a)!}color{blue}{x^a}color{green}{(1-x)^{b-a}}$$



              $$frac{color{blue}{b!}}{color{green}{a!}color{purple}{(b-a-1)!}}x=frac{color{blue}{b!}}{color{green}{(a-1)!}color{purple}{(b-a)!}}(1-x)$$



              $$frac{x}{a}=frac{1-x}{b-a} implies x(b-a)=a(1-x) implies bx-ax=a-ax implies bx = a$$






              share|cite|improve this answer
























                1












                1








                1






                It’s just really a sequence of simplifications.



                $$(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$$



                $$color{blue}{(a+1)}frac{b!}{color{blue}{(a+1)!}(b-(a+1))!}x^{a+1}(1-x)^{b-a-1}=color{green}{a}frac{b!}{color{green}{a!}(b-a)!}x^a(1-x)^{b-a}$$



                $$frac{b!}{a!(b-a-1)!}color{blue}{x^{a+1}}color{green}{(1-x)^{b-a-1}}=frac{b!}{(a-1)!(b-a)!}color{blue}{x^a}color{green}{(1-x)^{b-a}}$$



                $$frac{color{blue}{b!}}{color{green}{a!}color{purple}{(b-a-1)!}}x=frac{color{blue}{b!}}{color{green}{(a-1)!}color{purple}{(b-a)!}}(1-x)$$



                $$frac{x}{a}=frac{1-x}{b-a} implies x(b-a)=a(1-x) implies bx-ax=a-ax implies bx = a$$






                share|cite|improve this answer












                It’s just really a sequence of simplifications.



                $$(a+1){bchoose a+1}x^{a+1}(1-x)^{b-a-1}=a{bchoose a}x^a(1-x)^{b-a}$$



                $$color{blue}{(a+1)}frac{b!}{color{blue}{(a+1)!}(b-(a+1))!}x^{a+1}(1-x)^{b-a-1}=color{green}{a}frac{b!}{color{green}{a!}(b-a)!}x^a(1-x)^{b-a}$$



                $$frac{b!}{a!(b-a-1)!}color{blue}{x^{a+1}}color{green}{(1-x)^{b-a-1}}=frac{b!}{(a-1)!(b-a)!}color{blue}{x^a}color{green}{(1-x)^{b-a}}$$



                $$frac{color{blue}{b!}}{color{green}{a!}color{purple}{(b-a-1)!}}x=frac{color{blue}{b!}}{color{green}{(a-1)!}color{purple}{(b-a)!}}(1-x)$$



                $$frac{x}{a}=frac{1-x}{b-a} implies x(b-a)=a(1-x) implies bx-ax=a-ax implies bx = a$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 at 17:42









                KM101

                3,988417




                3,988417























                    2














                    Unless $x=0$ or $x=1$, $$frac{(a+1)b!}{(a+1)!(b-a-1)!}x^{a+1}(1-x)^{b-a-1}=frac{ab!}{a!(b-a)!}x^a(1-x)^{b-a}$$



                    simplifies to



                    $$(b-a)x=a(1-x).$$






                    share|cite|improve this answer





















                    • It also further simplifies to bx = a.
                      – Ofya
                      Nov 28 at 17:25










                    • Since we know b > a, we can say that 1>x>0 as well
                      – Ofya
                      Nov 28 at 17:25










                    • Actually, in the context of the problem earlier I showed that $a=bx$, so this solves my problem, thank you!
                      – I suck at Maths
                      Nov 28 at 17:36
















                    2














                    Unless $x=0$ or $x=1$, $$frac{(a+1)b!}{(a+1)!(b-a-1)!}x^{a+1}(1-x)^{b-a-1}=frac{ab!}{a!(b-a)!}x^a(1-x)^{b-a}$$



                    simplifies to



                    $$(b-a)x=a(1-x).$$






                    share|cite|improve this answer





















                    • It also further simplifies to bx = a.
                      – Ofya
                      Nov 28 at 17:25










                    • Since we know b > a, we can say that 1>x>0 as well
                      – Ofya
                      Nov 28 at 17:25










                    • Actually, in the context of the problem earlier I showed that $a=bx$, so this solves my problem, thank you!
                      – I suck at Maths
                      Nov 28 at 17:36














                    2












                    2








                    2






                    Unless $x=0$ or $x=1$, $$frac{(a+1)b!}{(a+1)!(b-a-1)!}x^{a+1}(1-x)^{b-a-1}=frac{ab!}{a!(b-a)!}x^a(1-x)^{b-a}$$



                    simplifies to



                    $$(b-a)x=a(1-x).$$






                    share|cite|improve this answer












                    Unless $x=0$ or $x=1$, $$frac{(a+1)b!}{(a+1)!(b-a-1)!}x^{a+1}(1-x)^{b-a-1}=frac{ab!}{a!(b-a)!}x^a(1-x)^{b-a}$$



                    simplifies to



                    $$(b-a)x=a(1-x).$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 28 at 17:24









                    Yves Daoust

                    124k671221




                    124k671221












                    • It also further simplifies to bx = a.
                      – Ofya
                      Nov 28 at 17:25










                    • Since we know b > a, we can say that 1>x>0 as well
                      – Ofya
                      Nov 28 at 17:25










                    • Actually, in the context of the problem earlier I showed that $a=bx$, so this solves my problem, thank you!
                      – I suck at Maths
                      Nov 28 at 17:36


















                    • It also further simplifies to bx = a.
                      – Ofya
                      Nov 28 at 17:25










                    • Since we know b > a, we can say that 1>x>0 as well
                      – Ofya
                      Nov 28 at 17:25










                    • Actually, in the context of the problem earlier I showed that $a=bx$, so this solves my problem, thank you!
                      – I suck at Maths
                      Nov 28 at 17:36
















                    It also further simplifies to bx = a.
                    – Ofya
                    Nov 28 at 17:25




                    It also further simplifies to bx = a.
                    – Ofya
                    Nov 28 at 17:25












                    Since we know b > a, we can say that 1>x>0 as well
                    – Ofya
                    Nov 28 at 17:25




                    Since we know b > a, we can say that 1>x>0 as well
                    – Ofya
                    Nov 28 at 17:25












                    Actually, in the context of the problem earlier I showed that $a=bx$, so this solves my problem, thank you!
                    – I suck at Maths
                    Nov 28 at 17:36




                    Actually, in the context of the problem earlier I showed that $a=bx$, so this solves my problem, thank you!
                    – I suck at Maths
                    Nov 28 at 17:36


















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