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In order to find the number of solutions to the equation $y_1 + y_2 + y_3 = 10$ where $y_i > 0, i \in$ {1,2,3}, then I let $y_i = x_i + 1$ with $x_i \in$ {0} $\cup \mathbb{N}$. Now I have $x_1 + x_2 + x_3 = 10 - 3 = 7$ and the solution is ${7 + 3 - 1 \choose 7}$ (if I'm not mistaken).

However, what if the restriction was that $ 2 <y_i < 8$ ? My way of thinking is that $y_i$ has now only 5 choices out of 11 but I don't know how to interpret this.

How should I think about such problem in order to come up with a solution, knowing how to solve the first problem described at the start of the question? I'm not looking for a direct solution to the problem but rather a way to think about this and understand the solution.

I read this but I don't get why $y_i = 6 - x_i$ (which would translate to $x_i = 5 - y_i$ to my problem [I think]).

alexandrosangeli
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  • If the restriction is $2<y_i<8$ I see only the solution $y_1=3;;y_2=3;;y_3=4$ and its two similar $(3,4,3);;(4,3,3)$ – Raffaele Feb 02 '21 at 16:12
  • Well this means that $x_i = 5 - y_i$ is the correct approach to this. But I still do not understand we do that. I understand that if the restriction was $y_i > 0$ then we do $x_i = y_i + 1$ so that every number is at least 1. But I don't get why we do what we do when the restriction is the opposite i.e. $y_i < N$. – alexandrosangeli Feb 02 '21 at 16:49

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