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?
Indeed. So this is $\lvert \{0,1,2,\dots,100\}\rvert = 101$ 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
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.