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?
$${{9}\choose{2}}-2\cdot\frac{{9}\choose{2}}{3}$$
 A: Since $10$ is not a multiple of $3$, it is not possible for all three numbers to be the same.
However, it is possible for two numbers of the same.  Since each number is positive, the repeated number must be $1$, $2$, $3$, or $4$.  There are $\binom{3}{2}$ ways to choose the locations of the repeated number.  Choosing the repeated number determines the third number.  There are 
$$4\binom{3}{2} = 12$$
such ordered positive integer solutions of the equation $a + b + c = 10$.  We count each unordered solution of this type three times, depending on where the single number is located.
That leaves $36 - 12 = 24$ solutions of the equation $a + b + c = 10$ in which all three numbers are distinct.  Since $a, b, c$ can be arranged in $3! = 6$ ways, we count each unordered solution of this type $6$ times.
Hence, the number of unordered positive integer solutions to the equation 
$a + b + c = 10$ is 
$$\frac{12}{3} + \frac{24}{6} = 4 + 4 = 8$$
As a check, we list them.
$$\{1, 2, 7\}, \{1, 3, 6\}, \{1, 4, 5\}, \{2, 3, 5\}, \{1, 1, 8\}, \{2, 2, 6\}, \{3, 3, 4\}, \{2, 4, 4\}$$
A: 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.
