<p>In how many ways can 14 identical toys be distributed among three boys so that each one gets at least one toy and no two boys get equal number of toys?</p>
Step-by-Step Solution
Key Concept: Transform the constraint 'each gets at least one' by giving 1 toy to each boy first, then distribute remaining 11 toys such that final counts are distinct. This reduces to finding ordered partitions of 11 into three distinct positive parts.
<p><strong>Step 1: Apply 'at least one' constraint</strong></p><p>Give 1 toy to each of the 3 boys (uses 3 toys). Remaining toys to distribute: 14 - 3 = 11</p><p><strong>Step 2: Set up the distinctness condition</strong></p><p>If boys receive x, y, z additional toys respectively, then total toys are (1+x), (1+y), (1+z) where x, y, z ≥ 0 and x + y + z = 11. We need 1+x, 1+y, 1+z to be all different, meaning x, y, z must be all different non-negative integers.</p><p><strong>Step 3: Find partitions of 11 into 3 distinct non-negative integers</strong></p><p>We need x ≠ y ≠ z ≠ x with x + y + z = 11 and x, y, z ≥ 0.</p><p>Unordered sets {x, y, z}:</p><p>• {0, 1, 10}, {0, 2, 9}, {0, 3, 8}, {0, 4, 7}, {0, 5, 6}</p><p>• {1, 2, 8}, {1, 3, 7}, {1, 4, 6}, {2, 3, 6}, {2, 4, 5}</p><p>Total: 10 unordered sets</p><p><strong>Step 4: Account for assignment to distinct boys</strong></p><p>Since the 3 boys are distinct, each unordered partition {a, b, c} can be assigned in 3! = 6 ways.</p><p>Total ways = 10 × 6 = 60</p><p>∴ Answer: <strong>60</strong></p>
Correct Answer: 60