Question

# how many solutions does the equation x1+x2+x3=13 have where x1,x2,and x3 are non negative integers less than 6

Algebra foundations

how many solutions does the equation $$\displaystyle{x}{1}+{x}{2}+{x}{3}={13}$$ have where $$x_1,x_2,\ and\ x_3$$ are non negative integers less than 6

2021-01-16

Without counting the symmetries, there are 2 ways you could do this. I will use $$\displaystyle{a}+{b}+{c}={13}.$$
The following solutions are to be associated to the pair (a,b,c):
$$\displaystyle{\left({5},{5},{3}\right)},{\left({4},{5},{4}\right)}$$ When you count the symmetries, each solution has 6 different arrangements and since there are 2 different solutions, you have a total of 12 solutions (symmetries included).
Here is why:
Since $$\displaystyle{0}≤{a},{b},{c}{<}{6},$$ it must be that $$\displaystyle{a}+{b}{<}{12}$$ and since no solutions will exist for $$\displaystyle{a}+{b}={11}$$, we consider instead that $$\displaystyle{a}+{b}\leq{10}$$. From here on, we continue cutting our possibilities to the following:
$$\displaystyle{a}+{b}={10},{c}={3}$$
$$\displaystyle{a}+{b}={9},{c}={4}$$
$$\displaystyle{a}+{b}={8},{c}={5}$$
The rest is just figuring out the values of aa and bb that satisfy each case, which is also easy.