Collections of Sets, Problem 4

Solution for (a)

Since we know that none of the three variables can be zero, three units of stuff are already accounted for. So we can rewrite the problem as

How many positive integer solutions are there for the equation \(a'+b'+c'=8\)?

where \(a'= a-1\), \(b' = b-1\), and \(c' = c-1\).

Using the formula for combinations with repetition, we have three types (two dividers) and 8 objects. So \({ {8+2} \choose 2 } = {{10} \choose 2}\) ways to do this.