How many non-negative integer solutions does (x+y+z=10) have if (x\leq3)?

Correct answer: (38)

Without the restriction on \(x\), the number of non-negative solutions of \(x+y+z=10\) is found by stars and bars: \(\binom{10+3-1}{3-1}=\binom{12}{2}=66\). We must remove the solutions where \(x\geq4\), because the required condition is \(x\leq3\). Let \(x'=x-4\), so \(x'\) is non-negative.

The changed equation is \(x'+y+z=6\), whose number of solutions is \(\binom{8}{2}=28\). Hence the desired number is \(66-28=38\). This counts every non-negative solution exactly once and removes precisely those with an excessive value of \(x\). Therefore option C is correct.