Why did we add 1 to each term?

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In how many ways can a manager of a chain of a supermarkets divide between two of its branches 100 units of one product, 200 of another, 150 of a third, and 11 of a fourth? [Notice divide means no branch is left without products]



Solution
(100+1)·(200+1)·(150+1)·(11+1)−2
Of the first product, the manager can assign k = 0, · · · , 100 units to one of the branches, and so for the other products. Since the products must be divided between the two branches, then we should not count the cases where all products are assigned to one branch.However, I don’t understand why we added 1 to each term?




probability statistics




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    up vote
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    down vote

    favorite












    In how many ways can a manager of a chain of a supermarkets divide between two of its branches 100 units of one product, 200 of another, 150 of a third, and 11 of a fourth? [Notice divide means no branch is left without products]



    Solution
    (100+1)·(200+1)·(150+1)·(11+1)−2
    Of the first product, the manager can assign k = 0, · · · , 100 units to one of the branches, and so for the other products. Since the products must be divided between the two branches, then we should not count the cases where all products are assigned to one branch.However, I don’t understand why we added 1 to each term?




    probability statistics




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      In how many ways can a manager of a chain of a supermarkets divide between two of its branches 100 units of one product, 200 of another, 150 of a third, and 11 of a fourth? [Notice divide means no branch is left without products]



      Solution
      (100+1)·(200+1)·(150+1)·(11+1)−2
      Of the first product, the manager can assign k = 0, · · · , 100 units to one of the branches, and so for the other products. Since the products must be divided between the two branches, then we should not count the cases where all products are assigned to one branch.However, I don’t understand why we added 1 to each term?




      probability statistics




      share|cite|improve this question











      In how many ways can a manager of a chain of a supermarkets divide between two of its branches 100 units of one product, 200 of another, 150 of a third, and 11 of a fourth? [Notice divide means no branch is left without products]



      Solution
      (100+1)·(200+1)·(150+1)·(11+1)−2
      Of the first product, the manager can assign k = 0, · · · , 100 units to one of the branches, and so for the other products. Since the products must be divided between the two branches, then we should not count the cases where all products are assigned to one branch.However, I don’t understand why we added 1 to each term?





      probability statistics





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      asked Jul 27 at 18:56









      Roy Rizk

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          1 Answer
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          "Of the first product, the manager can assign $k = 0, dots , 100$ units to one of the branches"




          Indeed. So this is $lvert 0,1,2,dots,100rvert = 101$ options for $k$ (the number of products of the first type put into the first branch can be zero).
          Same for the other 3 products. So you get
          $$
          (100+1)(200+1)(150+1)(11+1)
          $$
          possible assignments for the first branch (and, obviously, an assignment to the first branch fully determines the assignment to the second branch: the rest of the products).



          Now, there is the remaining $-2$. Why? Because




          divide means no branch is left without products




          so you must remove the two solutions corresponding to $0$, or all, products assigned to the first branch.



          Thus, overall, you obtain
          $$
          (100+1)(200+1)(150+1)(11+1)-2
          $$
          possible assignments.






          share|cite|improve this answer





















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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

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            up vote
            3
            down vote



            accepted











            "Of the first product, the manager can assign $k = 0, dots , 100$ units to one of the branches"




            Indeed. So this is $lvert 0,1,2,dots,100rvert = 101$ options for $k$ (the number of products of the first type put into the first branch can be zero).
            Same for the other 3 products. So you get
            $$
            (100+1)(200+1)(150+1)(11+1)
            $$
            possible assignments for the first branch (and, obviously, an assignment to the first branch fully determines the assignment to the second branch: the rest of the products).



            Now, there is the remaining $-2$. Why? Because




            divide means no branch is left without products




            so you must remove the two solutions corresponding to $0$, or all, products assigned to the first branch.



            Thus, overall, you obtain
            $$
            (100+1)(200+1)(150+1)(11+1)-2
            $$
            possible assignments.






            share|cite|improve this answer

























              up vote
              3
              down vote



              accepted











              "Of the first product, the manager can assign $k = 0, dots , 100$ units to one of the branches"




              Indeed. So this is $lvert 0,1,2,dots,100rvert = 101$ options for $k$ (the number of products of the first type put into the first branch can be zero).
              Same for the other 3 products. So you get
              $$
              (100+1)(200+1)(150+1)(11+1)
              $$
              possible assignments for the first branch (and, obviously, an assignment to the first branch fully determines the assignment to the second branch: the rest of the products).



              Now, there is the remaining $-2$. Why? Because




              divide means no branch is left without products




              so you must remove the two solutions corresponding to $0$, or all, products assigned to the first branch.



              Thus, overall, you obtain
              $$
              (100+1)(200+1)(150+1)(11+1)-2
              $$
              possible assignments.






              share|cite|improve this answer























                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted







                "Of the first product, the manager can assign $k = 0, dots , 100$ units to one of the branches"




                Indeed. So this is $lvert 0,1,2,dots,100rvert = 101$ options for $k$ (the number of products of the first type put into the first branch can be zero).
                Same for the other 3 products. So you get
                $$
                (100+1)(200+1)(150+1)(11+1)
                $$
                possible assignments for the first branch (and, obviously, an assignment to the first branch fully determines the assignment to the second branch: the rest of the products).



                Now, there is the remaining $-2$. Why? Because




                divide means no branch is left without products




                so you must remove the two solutions corresponding to $0$, or all, products assigned to the first branch.



                Thus, overall, you obtain
                $$
                (100+1)(200+1)(150+1)(11+1)-2
                $$
                possible assignments.






                share|cite|improve this answer














                "Of the first product, the manager can assign $k = 0, dots , 100$ units to one of the branches"




                Indeed. So this is $lvert 0,1,2,dots,100rvert = 101$ options for $k$ (the number of products of the first type put into the first branch can be zero).
                Same for the other 3 products. So you get
                $$
                (100+1)(200+1)(150+1)(11+1)
                $$
                possible assignments for the first branch (and, obviously, an assignment to the first branch fully determines the assignment to the second branch: the rest of the products).



                Now, there is the remaining $-2$. Why? Because




                divide means no branch is left without products




                so you must remove the two solutions corresponding to $0$, or all, products assigned to the first branch.



                Thus, overall, you obtain
                $$
                (100+1)(200+1)(150+1)(11+1)-2
                $$
                possible assignments.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 27 at 19:05









                Clement C.

                47k33682




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