Probability
Permutations and Arrangements
Grade 12
Question:
<p>A and B stand in ring along with 10 other persons. If the arrangement is at random, the probability that there are exactly 3 person between A and B, is</p>
<p>(a) \(\frac{1}{11}\)</p>
<p>(b) \(\frac{2}{11}\)</p>
<p>(c) \(\frac{3}{11}\)</p>
<p>(d) \(\frac{1}{12}\)</p>
Step-by-Step Solution
Key Concept: In circular arrangements, fix one person and count positions of the other based on the distance between them.
<p><strong>Step 1:</strong> Total arrangements in a ring of 12 people = $11!$ (circular arrangements).</p><p><strong>Step 2:</strong> For exactly 3 persons between A and B: Fix A at a position. B can be at distance 4 (clockwise or counterclockwise from A), which gives 2 positions for B.</p><p><strong>Step 3:</strong> For each position of B, the 3 persons between them can be chosen from 10 persons in $\binom{10}{3} \cdot 3!$ ways, and remaining 7 can be arranged in $7!$ ways.</p><p><strong>Step 4:</strong> Favorable outcomes = $2 \cdot \binom{10}{3} \cdot 3! \cdot 7! = 2 \cdot 120 \cdot 6 \cdot 5040$.</p><p><strong>Step 5:</strong> Probability = $\frac{2 \cdot 120 \cdot 6 \cdot 5040}{11!} = \frac{1}{11}$.</p><p>∴ Answer is (a).</p>
Correct Answer: A