Permutations & Combinations
Counting
Grade 11
Question:
<p>We have \(\lfloor 3(6+5+4+3) \rfloor = 6(18) = 108\). Since there are 3 ways for each group and 4 groups, as shown: 3 ways, 3 ways, 3 ways, 3 ways. The value of \(\lfloor 3(6+5+4+3) \rfloor\) is:</p>
<p>81</p>
<p>96</p>
<p>108</p>
<p>120</p>
Step-by-Step Solution
Key Concept: Recognize that when distributing identical items into distinct groups with constraints, the floor function collapses to exact integer calculation. Here, 3(6+5+4+3) = 3(18) = 54, and we must verify this matches the combinatorial structure of 3 choices across 4 independent groups.
<p><strong>Step 1:</strong> Calculate the sum inside the parentheses: 6 + 5 + 4 + 3 = 18</p><p><strong>Step 2:</strong> Multiply by the coefficient: 3 × 18 = 54</p><p><strong>Step 3:</strong> Apply the floor function: ⌊54⌋ = 54 (since 54 is already an integer)</p><p><strong>Step 4:</strong> Verify against the constraint structure: With 4 groups each having 3 independent choices, the total arrangements equal 3^4 = 81, but the given expression yields 54 as the relevant counting value for this specific configuration.</p><p>∴ Answer: <strong>54</strong> (Note: The value 108 appears to result from an error where the coefficient was doubled to 6 instead of 3)</p>
Correct Answer: C