<s>[INST] How many solutions does the equation $a + b + c = 10$ if $(1, 4, 5)$ and $(1, 5, 4)$ are not considered distinct? 
Question: What if we consider (1,4,5) and (1,5,4) as non-distinct possibilities, then what should we do?
$${{9}\choose{2}}-2\cdot\frac{{9}\choose{2}}{3}$$
 [/INST] Since solutions obtained by a permutation are considered the same we may assume $1\leq a\leq b\leq c$ to begin with. We therefore put
$$a=1+x_1,\quad b=a+x_2=1+x_1+x_2,\quad c=b+x_3=1+x_1+x_2+x_3\ ,$$
whereby  $x_i\geq0$ $\>(1\leq i\leq3)$ and
$a+b+c=3+3x_1+2x_2+x_3=10$, 
or $$3x_1+2x_2+x_3=7\ .$$
If $x_1=0$ then $2x_2+x_3=7$, which leads to the four solutions $(0,0,7)$, $(0,1,5)$, $(0,2,3)$, $(0,3,1)$.
If $x_1=1$ then $2x_2+x_3=4$, which leads to the three solutions $(1,0,4)$, $(1,1,2)$, $(1,2,0)$.
If $x_1=2$ we obtain the single solution $(2,0,1)$.
In terms of $a$ $b$, $c$  we get the $8$ solutions listed by N.F. Taussig. It remains unclear how your text arrived at $36$ solutions.
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