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For an equation,
a + b + c = D

where D is a natural number and a, b, c are variables, what is the quickest way to find the number of natural number solutions to this equation?

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closed as off-topic by Namaste, Thomas Andrews, Xam, Ethan Bolker, kingW3 Nov 11 '17 at 20:07

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    $\begingroup$ First, share with us one way you know to find the same. Then we can better understand what might be quicker. $\endgroup$ – Namaste Nov 11 '17 at 18:07
  • $\begingroup$ @amWhy That would be taking values from 1 to D - 2 for one of the variables and manually counting the possibilities for the the other two variables. For example. when c = 1, a + b = D - 1 , now I would count all possibilities for this, then take c = 2, and keep going until D - 2. $\endgroup$ – Nick Adams Nov 12 '17 at 7:46
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Stars and bars.

The problem can be reformulated in the following way:

Suppose we have $D+2$ places where we want to put two bars and $D$ stars. Thus the number of solutions is $D+2 \choose 2$.

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