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

Cem Hayes

Cem Hayes

Answered question

2021-01-15

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

Answer & Explanation

Demi-Leigh Barrera

Demi-Leigh Barrera

Skilled2021-01-16Added 97 answers

Without counting the symmetries, there are 2 ways you could do this. I will use a+b+c=13.
The following solutions are to be associated to the pair (a,b,c):
(5,5,3),(4,5,4) 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 0a,b,c<6, it must be that a+b<12 and since no solutions will exist for a+b=11, we consider instead that a+b10. From here on, we continue cutting our possibilities to the following:
a+b=10,c=3
a+b=9,c=4
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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