<p>In how many different ways may 12 things, 4 each of three varieties, be distributed equally among two persons?</p>
Step-by-Step Solution
Key Concept: The problem requires distributing 12 things (4 each of three varieties) equally among 2 persons, meaning each gets 6 things. Since items of the same variety are identical, we only count distinct distributions by how many of each variety each person receives.
<p><strong>Step 1:</strong> Recognize that each person must get exactly 6 items total. We need to distribute 4 items of variety A, 4 items of variety B, and 4 items of variety C between two persons.</p><p><strong>Step 2:</strong> For each variety, if person 1 gets <em>k</em> items, person 2 gets (4-<em>k</em>) items, where k ∈ {0,1,2,3,4}. This gives 5 choices per variety.</p><p><strong>Step 3:</strong> However, we need the constraint that person 1 gets exactly 6 items total. If person 1 gets a items of variety A, b items of variety B, and c items of variety C, then a + b + c = 6, where 0 ≤ a,b,c ≤ 4.</p><p><strong>Step 4:</strong> Find non-negative integer solutions to a + b + c = 6 with a,b,c ≤ 4:</p><p>• (4,2,0), (4,1,1), (4,0,2)</p><p>• (3,3,0), (3,2,1), (3,1,2), (3,0,3)</p><p>• (2,4,0), (2,3,1), (2,2,2), (2,1,3), (2,0,4)</p><p>• (1,4,1), (1,3,2), (1,2,3), (1,1,4)</p><p>• (0,4,2), (0,3,3), (0,2,4)</p><p><strong>Step 5:</strong> Counting all valid solutions: 3 + 4 + 5 + 4 + 3 = <strong>19</strong></p><p>∴ Answer: <strong>19</strong></p>
Correct Answer: 19