Why is $prod_{k=n}^{2n-3}left(2n-kright) = prod_{j=3}^{n}j$?












2












$begingroup$


I am trying to figure out these two equal expressions from my textbook:
$$prod_{k=n}^{2n-3}left(2n-kright) = prod _ {j=3} ^ {n}j$$



I have checked them and know that they apply, but what are the logical steps that lead me to the equation?






sequences-and-series discrete-mathematics







share|cite|improve this question











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    2












    $begingroup$


    I am trying to figure out these two equal expressions from my textbook:
    $$prod_{k=n}^{2n-3}left(2n-kright) = prod _ {j=3} ^ {n}j$$



    I have checked them and know that they apply, but what are the logical steps that lead me to the equation?






    sequences-and-series discrete-mathematics







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      I am trying to figure out these two equal expressions from my textbook:
      $$prod_{k=n}^{2n-3}left(2n-kright) = prod _ {j=3} ^ {n}j$$



      I have checked them and know that they apply, but what are the logical steps that lead me to the equation?






      sequences-and-series discrete-mathematics







      share|cite|improve this question











      $endgroup$




      I am trying to figure out these two equal expressions from my textbook:
      $$prod_{k=n}^{2n-3}left(2n-kright) = prod _ {j=3} ^ {n}j$$



      I have checked them and know that they apply, but what are the logical steps that lead me to the equation?






      sequences-and-series discrete-mathematics




      sequences-and-series discrete-mathematics






      share|cite|improve this question















      share|cite|improve this question













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      share|cite|improve this question








      edited Dec 19 '18 at 17:13







      Lucky

















      asked Dec 19 '18 at 17:00









      LuckyLucky

      1127




      1127






















          3 Answers
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          $begingroup$

          Let's write out the terms and see if we can observe a pattern.
          $$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
          Which simplifies to
          $$n(n-1)(n-2)(n-3)...(3)$$
          Which is equal to the desired product, since multiplication is commutative.






          share|cite|improve this answer









          $endgroup$





















            5












            $begingroup$

            Hint: Using a simple change variable:
            $$j = 2n - k$$



            Hence, its range is:
            $$ 2n - n = n geq j geq 2n-(2n-3) = 3$$






            share|cite|improve this answer









            $endgroup$





















              2












              $begingroup$

              begin{equation}
              prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
              end{equation}
              Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$






              share|cite|improve this answer









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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

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                3












                $begingroup$

                Let's write out the terms and see if we can observe a pattern.
                $$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
                Which simplifies to
                $$n(n-1)(n-2)(n-3)...(3)$$
                Which is equal to the desired product, since multiplication is commutative.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Let's write out the terms and see if we can observe a pattern.
                  $$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
                  Which simplifies to
                  $$n(n-1)(n-2)(n-3)...(3)$$
                  Which is equal to the desired product, since multiplication is commutative.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Let's write out the terms and see if we can observe a pattern.
                    $$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
                    Which simplifies to
                    $$n(n-1)(n-2)(n-3)...(3)$$
                    Which is equal to the desired product, since multiplication is commutative.






                    share|cite|improve this answer









                    $endgroup$



                    Let's write out the terms and see if we can observe a pattern.
                    $$(2n-n)(2n-(n+1))(2n-(n+2))(2n-(n+3))...(2n-(2n-3))$$
                    Which simplifies to
                    $$n(n-1)(n-2)(n-3)...(3)$$
                    Which is equal to the desired product, since multiplication is commutative.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 19 '18 at 17:07









                    user667user667

                    164117




                    164117























                        5












                        $begingroup$

                        Hint: Using a simple change variable:
                        $$j = 2n - k$$



                        Hence, its range is:
                        $$ 2n - n = n geq j geq 2n-(2n-3) = 3$$






                        share|cite|improve this answer









                        $endgroup$


















                          5












                          $begingroup$

                          Hint: Using a simple change variable:
                          $$j = 2n - k$$



                          Hence, its range is:
                          $$ 2n - n = n geq j geq 2n-(2n-3) = 3$$






                          share|cite|improve this answer









                          $endgroup$
















                            5












                            5








                            5





                            $begingroup$

                            Hint: Using a simple change variable:
                            $$j = 2n - k$$



                            Hence, its range is:
                            $$ 2n - n = n geq j geq 2n-(2n-3) = 3$$






                            share|cite|improve this answer









                            $endgroup$



                            Hint: Using a simple change variable:
                            $$j = 2n - k$$



                            Hence, its range is:
                            $$ 2n - n = n geq j geq 2n-(2n-3) = 3$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 19 '18 at 17:04









                            OmGOmG

                            2,502822




                            2,502822























                                2












                                $begingroup$

                                begin{equation}
                                prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
                                end{equation}
                                Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$






                                share|cite|improve this answer









                                $endgroup$


















                                  2












                                  $begingroup$

                                  begin{equation}
                                  prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
                                  end{equation}
                                  Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    2












                                    2








                                    2





                                    $begingroup$

                                    begin{equation}
                                    prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
                                    end{equation}
                                    Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$






                                    share|cite|improve this answer









                                    $endgroup$



                                    begin{equation}
                                    prod_{k=n}^{2n-3}left(2n-kright) = (2n - n)(2n - (n+1))ldots (2n - (2n-3)) = (n)(n-1)(n-2) ldots 3 = prod_{j=3}^n j
                                    end{equation}
                                    Another way is to do a change of variable, take $j = 2n- k$, then the extremities go from $j_1$ to $j_2$ where $j_1 = 2n - n = n$ and $j_2 = 2n - (2n - 3) = 3$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 19 '18 at 17:14









                                    Ahmad BazziAhmad Bazzi

                                    8,2962824




                                    8,2962824






























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