Binomial coefficient formula reduction












0












$begingroup$


I think I need to use $(x+y)^n = sum_{k=0}^n {n choose k} x^{n-k} y^k$, but I can't figure out how. I am trying to see why



$$frac{y!}{x!(y-x)!} lambda^x mu^{y-x} / sum_{x,r:x+r=y}frac{y!}{x! r!} lambda^x mu^r$$



is the same as



$${y choose x} left(frac{lambda}{lambda + mu} right)^x left( frac{mu}{lambda + mu} right)^{y-x}$$



But can't see why






statistics binomial-coefficients







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    I think I need to use $(x+y)^n = sum_{k=0}^n {n choose k} x^{n-k} y^k$, but I can't figure out how. I am trying to see why



    $$frac{y!}{x!(y-x)!} lambda^x mu^{y-x} / sum_{x,r:x+r=y}frac{y!}{x! r!} lambda^x mu^r$$



    is the same as



    $${y choose x} left(frac{lambda}{lambda + mu} right)^x left( frac{mu}{lambda + mu} right)^{y-x}$$



    But can't see why






    statistics binomial-coefficients







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I think I need to use $(x+y)^n = sum_{k=0}^n {n choose k} x^{n-k} y^k$, but I can't figure out how. I am trying to see why



      $$frac{y!}{x!(y-x)!} lambda^x mu^{y-x} / sum_{x,r:x+r=y}frac{y!}{x! r!} lambda^x mu^r$$



      is the same as



      $${y choose x} left(frac{lambda}{lambda + mu} right)^x left( frac{mu}{lambda + mu} right)^{y-x}$$



      But can't see why






      statistics binomial-coefficients







      share|cite|improve this question









      $endgroup$




      I think I need to use $(x+y)^n = sum_{k=0}^n {n choose k} x^{n-k} y^k$, but I can't figure out how. I am trying to see why



      $$frac{y!}{x!(y-x)!} lambda^x mu^{y-x} / sum_{x,r:x+r=y}frac{y!}{x! r!} lambda^x mu^r$$



      is the same as



      $${y choose x} left(frac{lambda}{lambda + mu} right)^x left( frac{mu}{lambda + mu} right)^{y-x}$$



      But can't see why






      statistics binomial-coefficients




      statistics binomial-coefficients






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 21:57









      i squared - Keep it Reali squared - Keep it Real

      1,62511127




      1,62511127






















          3 Answers
          3






          active

          oldest

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          1












          $begingroup$

          The numerator of your LHS is by definition of ${n choose k}$ equal to $${y choose x} lambda^x mu^{y-x}$$
          The denominator can be rewritten as:
          $$sum_{r=0}^y {y choose r} mu^r lambda^{y-r} = (mu+lambda)^y = (mu+lambda)^x (mu+lambda)^{y-x}$$



          Now if you take a ratio of the simplified numerator and denominator, you will get the desired answer,






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            It seems that the index r starts at $r=0$ and the maximum value of r is if $x=0$. That means $r=y-x=y-0=y$.



            $$sum_{r=0}^yfrac{y!}{x! r!} lambda^x mu^r overset{substitution}{=}sum_{r=0}^yfrac{y!}{(y-r)! r!}lambda^{y-r} mu^r$$



            Now apply the binomial theorem.






            share|cite|improve this answer











            $endgroup$





















              0












              $begingroup$

              If you match the two expressions and cancel terms, you find that you need to show
              $$(lambda + mu)^y = sum_{x,r : x+r = y} frac{y!}{x! r!} lambda^x mu^r.$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                how so, if in the second expression, the two denominators are: $(lambda + mu)^x$ and $(lambda + mu)^{y-x}$
                $endgroup$
                – i squared - Keep it Real
                Jan 8 at 22:26












              • $begingroup$
                which would mean that $(lambda + mu)^x(lambda + mu)^{y-x} = sum_{x,r:x+r=y} frac{y!}{x!r!}lambda^x mu^r$
                $endgroup$
                – i squared - Keep it Real
                Jan 8 at 22:27










              • $begingroup$
                which is $(lambda + mu)^y$...
                $endgroup$
                – i squared - Keep it Real
                Jan 8 at 22:28










              • $begingroup$
                can't show it..
                $endgroup$
                – i squared - Keep it Real
                Jan 8 at 22:37










              • $begingroup$
                it is intuitive that that is the result because we are finding all the combinations that sum up to $y$. The crux must be in re-writing the $x,r:x+r=y$ intelligently
                $endgroup$
                – i squared - Keep it Real
                Jan 8 at 22:42












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              3 Answers
              3






              active

              oldest

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              3 Answers
              3






              active

              oldest

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              active

              oldest

              votes






              active

              oldest

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              1












              $begingroup$

              The numerator of your LHS is by definition of ${n choose k}$ equal to $${y choose x} lambda^x mu^{y-x}$$
              The denominator can be rewritten as:
              $$sum_{r=0}^y {y choose r} mu^r lambda^{y-r} = (mu+lambda)^y = (mu+lambda)^x (mu+lambda)^{y-x}$$



              Now if you take a ratio of the simplified numerator and denominator, you will get the desired answer,






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The numerator of your LHS is by definition of ${n choose k}$ equal to $${y choose x} lambda^x mu^{y-x}$$
                The denominator can be rewritten as:
                $$sum_{r=0}^y {y choose r} mu^r lambda^{y-r} = (mu+lambda)^y = (mu+lambda)^x (mu+lambda)^{y-x}$$



                Now if you take a ratio of the simplified numerator and denominator, you will get the desired answer,






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The numerator of your LHS is by definition of ${n choose k}$ equal to $${y choose x} lambda^x mu^{y-x}$$
                  The denominator can be rewritten as:
                  $$sum_{r=0}^y {y choose r} mu^r lambda^{y-r} = (mu+lambda)^y = (mu+lambda)^x (mu+lambda)^{y-x}$$



                  Now if you take a ratio of the simplified numerator and denominator, you will get the desired answer,






                  share|cite|improve this answer









                  $endgroup$



                  The numerator of your LHS is by definition of ${n choose k}$ equal to $${y choose x} lambda^x mu^{y-x}$$
                  The denominator can be rewritten as:
                  $$sum_{r=0}^y {y choose r} mu^r lambda^{y-r} = (mu+lambda)^y = (mu+lambda)^x (mu+lambda)^{y-x}$$



                  Now if you take a ratio of the simplified numerator and denominator, you will get the desired answer,







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 23:33









                  Aditya DuaAditya Dua

                  1,15418




                  1,15418























                      1












                      $begingroup$

                      It seems that the index r starts at $r=0$ and the maximum value of r is if $x=0$. That means $r=y-x=y-0=y$.



                      $$sum_{r=0}^yfrac{y!}{x! r!} lambda^x mu^r overset{substitution}{=}sum_{r=0}^yfrac{y!}{(y-r)! r!}lambda^{y-r} mu^r$$



                      Now apply the binomial theorem.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        It seems that the index r starts at $r=0$ and the maximum value of r is if $x=0$. That means $r=y-x=y-0=y$.



                        $$sum_{r=0}^yfrac{y!}{x! r!} lambda^x mu^r overset{substitution}{=}sum_{r=0}^yfrac{y!}{(y-r)! r!}lambda^{y-r} mu^r$$



                        Now apply the binomial theorem.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          It seems that the index r starts at $r=0$ and the maximum value of r is if $x=0$. That means $r=y-x=y-0=y$.



                          $$sum_{r=0}^yfrac{y!}{x! r!} lambda^x mu^r overset{substitution}{=}sum_{r=0}^yfrac{y!}{(y-r)! r!}lambda^{y-r} mu^r$$



                          Now apply the binomial theorem.






                          share|cite|improve this answer











                          $endgroup$



                          It seems that the index r starts at $r=0$ and the maximum value of r is if $x=0$. That means $r=y-x=y-0=y$.



                          $$sum_{r=0}^yfrac{y!}{x! r!} lambda^x mu^r overset{substitution}{=}sum_{r=0}^yfrac{y!}{(y-r)! r!}lambda^{y-r} mu^r$$



                          Now apply the binomial theorem.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 8 at 23:43

























                          answered Jan 8 at 23:34









                          callculuscallculus

                          18.7k31428




                          18.7k31428























                              0












                              $begingroup$

                              If you match the two expressions and cancel terms, you find that you need to show
                              $$(lambda + mu)^y = sum_{x,r : x+r = y} frac{y!}{x! r!} lambda^x mu^r.$$






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                how so, if in the second expression, the two denominators are: $(lambda + mu)^x$ and $(lambda + mu)^{y-x}$
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:26












                              • $begingroup$
                                which would mean that $(lambda + mu)^x(lambda + mu)^{y-x} = sum_{x,r:x+r=y} frac{y!}{x!r!}lambda^x mu^r$
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:27










                              • $begingroup$
                                which is $(lambda + mu)^y$...
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:28










                              • $begingroup$
                                can't show it..
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:37










                              • $begingroup$
                                it is intuitive that that is the result because we are finding all the combinations that sum up to $y$. The crux must be in re-writing the $x,r:x+r=y$ intelligently
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:42
















                              0












                              $begingroup$

                              If you match the two expressions and cancel terms, you find that you need to show
                              $$(lambda + mu)^y = sum_{x,r : x+r = y} frac{y!}{x! r!} lambda^x mu^r.$$






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                how so, if in the second expression, the two denominators are: $(lambda + mu)^x$ and $(lambda + mu)^{y-x}$
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:26












                              • $begingroup$
                                which would mean that $(lambda + mu)^x(lambda + mu)^{y-x} = sum_{x,r:x+r=y} frac{y!}{x!r!}lambda^x mu^r$
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:27










                              • $begingroup$
                                which is $(lambda + mu)^y$...
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:28










                              • $begingroup$
                                can't show it..
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:37










                              • $begingroup$
                                it is intuitive that that is the result because we are finding all the combinations that sum up to $y$. The crux must be in re-writing the $x,r:x+r=y$ intelligently
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:42














                              0












                              0








                              0





                              $begingroup$

                              If you match the two expressions and cancel terms, you find that you need to show
                              $$(lambda + mu)^y = sum_{x,r : x+r = y} frac{y!}{x! r!} lambda^x mu^r.$$






                              share|cite|improve this answer









                              $endgroup$



                              If you match the two expressions and cancel terms, you find that you need to show
                              $$(lambda + mu)^y = sum_{x,r : x+r = y} frac{y!}{x! r!} lambda^x mu^r.$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 8 at 22:00









                              angryavianangryavian

                              42.5k23481




                              42.5k23481












                              • $begingroup$
                                how so, if in the second expression, the two denominators are: $(lambda + mu)^x$ and $(lambda + mu)^{y-x}$
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:26












                              • $begingroup$
                                which would mean that $(lambda + mu)^x(lambda + mu)^{y-x} = sum_{x,r:x+r=y} frac{y!}{x!r!}lambda^x mu^r$
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:27










                              • $begingroup$
                                which is $(lambda + mu)^y$...
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:28










                              • $begingroup$
                                can't show it..
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:37










                              • $begingroup$
                                it is intuitive that that is the result because we are finding all the combinations that sum up to $y$. The crux must be in re-writing the $x,r:x+r=y$ intelligently
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:42


















                              • $begingroup$
                                how so, if in the second expression, the two denominators are: $(lambda + mu)^x$ and $(lambda + mu)^{y-x}$
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:26












                              • $begingroup$
                                which would mean that $(lambda + mu)^x(lambda + mu)^{y-x} = sum_{x,r:x+r=y} frac{y!}{x!r!}lambda^x mu^r$
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:27










                              • $begingroup$
                                which is $(lambda + mu)^y$...
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:28










                              • $begingroup$
                                can't show it..
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:37










                              • $begingroup$
                                it is intuitive that that is the result because we are finding all the combinations that sum up to $y$. The crux must be in re-writing the $x,r:x+r=y$ intelligently
                                $endgroup$
                                – i squared - Keep it Real
                                Jan 8 at 22:42
















                              $begingroup$
                              how so, if in the second expression, the two denominators are: $(lambda + mu)^x$ and $(lambda + mu)^{y-x}$
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:26






                              $begingroup$
                              how so, if in the second expression, the two denominators are: $(lambda + mu)^x$ and $(lambda + mu)^{y-x}$
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:26














                              $begingroup$
                              which would mean that $(lambda + mu)^x(lambda + mu)^{y-x} = sum_{x,r:x+r=y} frac{y!}{x!r!}lambda^x mu^r$
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:27




                              $begingroup$
                              which would mean that $(lambda + mu)^x(lambda + mu)^{y-x} = sum_{x,r:x+r=y} frac{y!}{x!r!}lambda^x mu^r$
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:27












                              $begingroup$
                              which is $(lambda + mu)^y$...
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:28




                              $begingroup$
                              which is $(lambda + mu)^y$...
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:28












                              $begingroup$
                              can't show it..
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:37




                              $begingroup$
                              can't show it..
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:37












                              $begingroup$
                              it is intuitive that that is the result because we are finding all the combinations that sum up to $y$. The crux must be in re-writing the $x,r:x+r=y$ intelligently
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:42




                              $begingroup$
                              it is intuitive that that is the result because we are finding all the combinations that sum up to $y$. The crux must be in re-writing the $x,r:x+r=y$ intelligently
                              $endgroup$
                              – i squared - Keep it Real
                              Jan 8 at 22:42


















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