Finite Mathematics Quiz: Combinations With Repetition
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Combinations With RepetitionQuestion 1 of 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?

6666 combinations
7878 combinations
8484 combinations
9696 combinations
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Finite Mathematics Quiz

Finite Mathematics Quiz: Combinations With Repetition

Practice Combinations With Repetition in Finite Mathematics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Combinations With Repetition, giving you a quick way to practice the rules, question types, and explanations that matter most for Finite Mathematics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 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?

  1. 6666 combinations (correct answer)
  2. 7878 combinations
  3. 8484 combinations
  4. 9696 combinations
Explanation: Let a,b,ca, b, c be cases from suppliers A, B, C respectively. We need a+b+c=25a + b + c = 25 where a≥5a ≥ 5, bb is even, and a+c≤20a + c ≤ 20. Since a+b+c=25a + b + c = 25 and a+c≤20a + c ≤ 20, we have b≥5b ≥ 5. Since bb must be even and b≥5b ≥ 5, we have b∈{6,8,10,12,14,16,18,20,22,24}b ∈ \{6, 8, 10, 12, 14, 16, 18, 20, 22, 24\}. For each valid bb, we need a+c=25−ba + c = 25 - b where a≥5a ≥ 5, c≥0c ≥ 0, and a+c≤20a + c ≤ 20. This gives us a+c=min⁡(25−b,20)a + c = \min(25 - b, 20). For b=6b = 6: a+c=19a + c = 19, with 5≤a≤195 ≤ a ≤ 19, giving 15 solutions. For b=8b = 8: a+c=17a + c = 17, with 5≤a≤175 ≤ a ≤ 17, giving 13 solutions. For b=10b = 10: a+c=15a + c = 15, with 5≤a≤155 ≤ a ≤ 15, giving 11 solutions. For b=12b = 12: a+c=13a + c = 13, with 5≤a≤135 ≤ a ≤ 13, giving 9 solutions. For b=14b = 14: a+c=11a + c = 11, with 5≤a≤115 ≤ a ≤ 11, giving 7 solutions. For b=16b = 16: a+c=9a + c = 9, with 5≤a≤95 ≤ a ≤ 9, giving 5 solutions. For b=18b = 18: a+c=7a + c = 7, with 5≤a≤75 ≤ a ≤ 7, giving 3 solutions. For b=20b = 20: a+c=5a + c = 5, with a=5,c=0a = 5, c = 0, giving 1 solution. For b≥22b ≥ 22: a+c≤3a + c ≤ 3, but a≥5a ≥ 5, so no solutions. Total: 15+13+11+9+7+5+3+1=6615 + 13 + 11 + 9 + 7 + 5 + 3 + 1 = 66.