Combinations with Repetition - Finite Mathematics

1. A restaurant manager is planning to order exactly 25 cases of beverages from 3 suppliers: Supplier A (soda), Supplier B (juice), and Supplier C (water). Due to storage constraints, orders from Supplier A must be at least 5 cases, orders from Supplier B must be even numbers of cases (including 0), and the total order from Suppliers A and C combined cannot exceed 20 cases. How many different ordering combinations are possible?
   Answer: $66$ combinations Let $a, b, c$ be cases from suppliers A, B, C respectively. We need $a + b + c = 25$ where $a ≥ 5$, $b$ is even, and $a + c ≤ 20$. Since $a + b + c = 25$ and $a + c ≤ 20$, we have $b ≥ 5$. Since $b$ must be even and $b ≥ 5$, we have $b ∈ \{6, 8, 10, 12, 14, 16, 18, 20, 22, 24\}$. For each valid $b$, we need $a + c = 25 - b$ where $a ≥ 5$, $c ≥ 0$, and $a + c ≤ 20$. This gives us $a + c = \min(25 - b, 20)$. For $b = 6$: $a + c = 19$, with $5 ≤ a ≤ 19$, giving 15 solutions. For $b = 8$: $a + c = 17$, with $5 ≤ a ≤ 17$, giving 13 solutions. For $b = 10$: $a + c = 15$, with $5 ≤ a ≤ 15$, giving 11 solutions. For $b = 12$: $a + c = 13$, with $5 ≤ a ≤ 13$, giving 9 solutions. For $b = 14$: $a + c = 11$, with $5 ≤ a ≤ 11$, giving 7 solutions. For $b = 16$: $a + c = 9$, with $5 ≤ a ≤ 9$, giving 5 solutions. For $b = 18$: $a + c = 7$, with $5 ≤ a ≤ 7$, giving 3 solutions. For $b = 20$: $a + c = 5$, with $a = 5, c = 0$, giving 1 solution. For $b ≥ 22$: $a + c ≤ 3$, but $a ≥ 5$, so no solutions. Total: $15 + 13 + 11 + 9 + 7 + 5 + 3 + 1 = 66$.

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