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Number of non negative integral solutions for $a + b + c = n$
Where $n$ is a positive integer are
$$\binom{n + 3 - 1}{3 - 1}$$

But if a condition is there $a > b > c$
Is there any direct method by which we can find out required number of solutions.

I believe that we should multiply original number of solutions by $1 / 4$ as there are four following cases possible $a = b > c$
$a = b = c$
$a > b = c$
$a > b > c$

  • 4
    Dividing by 4 would be appropriate only if the cases were equal in number and if you already had $a \ge b \ge c$. When $n$ is large, there should be many more solutions with $a,b,c$ all different than solutions with some equal. There are also cases with $c \gt b \gt a$, for example. Better to find the number of solutions with $a,b,c$ all different. Then you can divide by $6$ because there are that many orders of $a,b,c$2012-07-04
  • 0
    What if I give you a equation a + b + c = n where a >= b > = c2012-07-04
  • 0
    Is there any method to find number of solutions to original question.2012-07-04
  • 0
    I just removed the linear-algebra tag, it doesn't seem to be relevant.2012-07-04

2 Answers 2