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
ANSWERED
asked 2021-01-15

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

Answers (1)

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.

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