Permutations & Combinations
Distribution of distinct objects
Grade 11
Question:
<p>The number of ways to give 16 different things to three persons \(A\), \(B\), \(C\) so that \(B\) gets one more than \(A\) and \(C\) gets two more than \(B\) is</p>
<p>\(\dfrac{16!}{4!5!7!}\)</p>
<p>\(4! \cdot 5! \cdot 7!\)</p>
<p>\(\dfrac{16!}{3!5!8!}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Set up equations for the distribution: if A gets 'a' things, then B gets 'a+1' and C gets 'a+3', where a + (a+1) + (a+3) = 16. Solve for a, then use multinomial coefficients to count ways.
<p><strong>Step 1:</strong> Let A get <em>a</em> things. Then B gets <em>a</em> + 1 things, and C gets (<em>a</em> + 1) + 2 = <em>a</em> + 3 things.</p><p><strong>Step 2:</strong> Total things constraint: <em>a</em> + (<em>a</em> + 1) + (<em>a</em> + 3) = 16 → 3<em>a</em> + 4 = 16 → <em>a</em> = 4</p><p><strong>Step 3:</strong> So A gets 4, B gets 5, and C gets 7 things.</p><p><strong>Step 4:</strong> Since all 16 things are different, the number of ways to distribute them is the multinomial coefficient:</p><p>$$\binom{16}{4, 5, 7} = \frac{16!}{4! \cdot 5! \cdot 7!}$$</p><p><strong>Step 5:</strong> Calculate: $$\frac{16!}{4! \cdot 5! \cdot 7!} = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}{4! \times 5!} = 168168000$$</p><p>∴ Answer: <strong>168168000</strong> or <strong>\(\binom{16}{4,5,7}\)</strong></p>
Correct Answer: A