Split $151$ cakes amongst $3$ people












2












$begingroup$


$151$ cakes shall be split amongst $3$ people under the condition that no one must have more than $75$. How many combinations are possible?



The solution is $${binom {3+74-1}{3-1}}$$



which is exactly a combination with replacement.



I don't understand why the number of cakes can be reduced to $74$. There must be some kind of bijectivity between splitting $151$ cakes and the $74$ I don't see.






combinatorics







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    $151$ cakes shall be split amongst $3$ people under the condition that no one must have more than $75$. How many combinations are possible?



    The solution is $${binom {3+74-1}{3-1}}$$



    which is exactly a combination with replacement.



    I don't understand why the number of cakes can be reduced to $74$. There must be some kind of bijectivity between splitting $151$ cakes and the $74$ I don't see.






    combinatorics







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      $151$ cakes shall be split amongst $3$ people under the condition that no one must have more than $75$. How many combinations are possible?



      The solution is $${binom {3+74-1}{3-1}}$$



      which is exactly a combination with replacement.



      I don't understand why the number of cakes can be reduced to $74$. There must be some kind of bijectivity between splitting $151$ cakes and the $74$ I don't see.






      combinatorics







      share|cite|improve this question











      $endgroup$




      $151$ cakes shall be split amongst $3$ people under the condition that no one must have more than $75$. How many combinations are possible?



      The solution is $${binom {3+74-1}{3-1}}$$



      which is exactly a combination with replacement.



      I don't understand why the number of cakes can be reduced to $74$. There must be some kind of bijectivity between splitting $151$ cakes and the $74$ I don't see.






      combinatorics




      combinatorics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 5 hours ago









      Asaf Karagila♦

      306k33437768




      306k33437768










      asked 6 hours ago









      EpsilonDeltaEpsilonDelta

      6921615




      6921615






















          2 Answers
          2






          active

          oldest

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          5












          $begingroup$

          The solutions of the system :
          $$
          x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
          $$



          is in one-one correspondence with the solutions of the system :
          $$
          (75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
          $$



          Which after the change of variable $y_i = 75-x_i$ becomes (since $color{blue}{75times3 - 151 = 74}$):
          $$
          y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
          $$
          which by non-negativity has the same solutions as :
          $$
          y_1+y_2+y_3 = 74 : 0 leq y_i forall i
          $$



          Now, play the stars-and-bars game to find the number of solutions of this equation.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom{75}{2}!!!right)=binom{75+2-1}{2}=binom{74+3-1}{3-1}$$



            This requires some explanation: There is only one way of combining the two numbers to give three numbers less than 75. For example, if you choose 21 and 12, then think of 75 as being split into 21+54 and 12+63; the only way to combine them to give three acceptable numbers is as 54+33+63. In general, one must combine the two numbers that are smaller than half 75.






            share|cite|improve this answer











            $endgroup$













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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

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              5












              $begingroup$

              The solutions of the system :
              $$
              x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
              $$



              is in one-one correspondence with the solutions of the system :
              $$
              (75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
              $$



              Which after the change of variable $y_i = 75-x_i$ becomes (since $color{blue}{75times3 - 151 = 74}$):
              $$
              y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
              $$
              which by non-negativity has the same solutions as :
              $$
              y_1+y_2+y_3 = 74 : 0 leq y_i forall i
              $$



              Now, play the stars-and-bars game to find the number of solutions of this equation.






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                The solutions of the system :
                $$
                x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
                $$



                is in one-one correspondence with the solutions of the system :
                $$
                (75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
                $$



                Which after the change of variable $y_i = 75-x_i$ becomes (since $color{blue}{75times3 - 151 = 74}$):
                $$
                y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
                $$
                which by non-negativity has the same solutions as :
                $$
                y_1+y_2+y_3 = 74 : 0 leq y_i forall i
                $$



                Now, play the stars-and-bars game to find the number of solutions of this equation.






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  The solutions of the system :
                  $$
                  x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
                  $$



                  is in one-one correspondence with the solutions of the system :
                  $$
                  (75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
                  $$



                  Which after the change of variable $y_i = 75-x_i$ becomes (since $color{blue}{75times3 - 151 = 74}$):
                  $$
                  y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
                  $$
                  which by non-negativity has the same solutions as :
                  $$
                  y_1+y_2+y_3 = 74 : 0 leq y_i forall i
                  $$



                  Now, play the stars-and-bars game to find the number of solutions of this equation.






                  share|cite|improve this answer









                  $endgroup$



                  The solutions of the system :
                  $$
                  x_1 + x_2 + x_3 = 151 : 0 leq x_1,x_2,x_3 leq 75 \
                  $$



                  is in one-one correspondence with the solutions of the system :
                  $$
                  (75 - x_1) + (75 - x_2) + (75 - x_3) = 3 times 75 - 151 : 0 leq x_1,x_2,x_3 leq 75
                  $$



                  Which after the change of variable $y_i = 75-x_i$ becomes (since $color{blue}{75times3 - 151 = 74}$):
                  $$
                  y_1+y_2+y_3 = 74 : 0 leq y_1,y_2,y_3 leq 75
                  $$
                  which by non-negativity has the same solutions as :
                  $$
                  y_1+y_2+y_3 = 74 : 0 leq y_i forall i
                  $$



                  Now, play the stars-and-bars game to find the number of solutions of this equation.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 6 hours ago









                  астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

                  39.1k33477




                  39.1k33477























                      1












                      $begingroup$

                      The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom{75}{2}!!!right)=binom{75+2-1}{2}=binom{74+3-1}{3-1}$$



                      This requires some explanation: There is only one way of combining the two numbers to give three numbers less than 75. For example, if you choose 21 and 12, then think of 75 as being split into 21+54 and 12+63; the only way to combine them to give three acceptable numbers is as 54+33+63. In general, one must combine the two numbers that are smaller than half 75.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom{75}{2}!!!right)=binom{75+2-1}{2}=binom{74+3-1}{3-1}$$



                        This requires some explanation: There is only one way of combining the two numbers to give three numbers less than 75. For example, if you choose 21 and 12, then think of 75 as being split into 21+54 and 12+63; the only way to combine them to give three acceptable numbers is as 54+33+63. In general, one must combine the two numbers that are smaller than half 75.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom{75}{2}!!!right)=binom{75+2-1}{2}=binom{74+3-1}{3-1}$$



                          This requires some explanation: There is only one way of combining the two numbers to give three numbers less than 75. For example, if you choose 21 and 12, then think of 75 as being split into 21+54 and 12+63; the only way to combine them to give three acceptable numbers is as 54+33+63. In general, one must combine the two numbers that are smaller than half 75.






                          share|cite|improve this answer











                          $endgroup$



                          The number of possibilities is like choosing two numbers from 75 (with replacement). Hence it is $$left(!!!binom{75}{2}!!!right)=binom{75+2-1}{2}=binom{74+3-1}{3-1}$$



                          This requires some explanation: There is only one way of combining the two numbers to give three numbers less than 75. For example, if you choose 21 and 12, then think of 75 as being split into 21+54 and 12+63; the only way to combine them to give three acceptable numbers is as 54+33+63. In general, one must combine the two numbers that are smaller than half 75.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 2 hours ago

























                          answered 6 hours ago









                          ChrystomathChrystomath

                          1,343512




                          1,343512






























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