Find the number of solutions.

Question

Find the number of solutions $(x,y,z)$ of the equation $x+y+z=10$,where each of $x$,$y$,and $z$ is a positive integer.

Answer 1

Given $x+y+z=10\;,$ Where $x,y,z\geq 1$

Which is equivalent to distribute $10$ chocolates into $3$ students, such that at least get one chocolates, which can be done as

$$\bf{\underbrace{**}_{2\; chocolates}|\underbrace{***}_{3\; chocolates}|\underbrace{*****}_{5\; chocolates}}$$ as one example

Means arranging $2$ bars in $9$ gap(In between two square)

So total number of ways $$\binom{9}{2} = 36$$

Answer 2

hint: Let $x = 1+m, y = 1+n, z = 1 + p \Rightarrow m+n+p = 7$. This is well-known star-and-bar method for counting number of non-negative integer solutions of an equation like this one. Can you continue?

Answer 3

Hint:

If $x=1$, $y+z=9$, and there are $8$ ways to achieve that (from $1+8$ to $8+1$).

If $x=2$, $y+z=8$, and there are $7$ ways to achieve that (from $1+7$ to $7+1$).

...

If $x=8$, $y+z=2$, and there is $1$ way to achieve that ($1+1$).