Question

The number of positive integral solutions of $10 \leq y_1 + y_2 + y_3 < 15$ is equal to

• A. 356
• B. 355
• C. 306
• D. 280

Answer 1

Answer: (D)

280

Answer 2

The inequality $10 \leq y_1 + y_2 + y_3 < 15$ means the sum $S = y_1 + y_2 + y_3$ can take integer values $10, 11, 12, 13, 14$. The number of positive integral solutions for $y_1 + y_2 + y_3 = k$ is given by the stars and bars formula: $\binom{k-1}{3-1} = \binom{k-1}{2}$. We need to sum this for $k=10, 11, 12, 13, 14$. Total solutions = $\sum_{k=10}^{14} \binom{k-1}{2} = \binom{9}{2} + \binom{10}{2} + \binom{11}{2} + \binom{12}{2} + \binom{13}{2}$. Using the Hockey-stick identity $\sum_{i=r}^n \binom{i}{r} = \binom{n+1}{r+1}$, we can rewrite the sum as $\binom{14}{3} - \binom{9}{3} = 364 - 84 = 280$.