Probability
Derangements and combinatorial probability
Grade 12
Question:
<p>There are six students \(S_1, S_2, S_3, S_4, S_5\) and \(S_6\) and for them there are six seats \(R_1, R_2, R_3, R_4, R_5\) and \(R_6\) arranged in a row, where the initially seat \(R_i\) is allotted to the student \(S_i\), \(i = 1, 2, 3, 4, 5, 6\). But on a day all are allotted seats randomly.</p><p>Which of the following statements are correct?</p><p>(a) The probability that \(S_4\) gets allotted seat \(R_4\) and none of the remaining students gets the seat previously allotted to them, is \(\dfrac{11}{180}\).</p><p>(b) The probability that \(S_5\) gets allotted seat \(R_5\) and none of the remaining gets the seat previously allotted to them is \(\dfrac{1}{6}\).</p><p>(c) The probability that even numbered students are seating at even numbered seat and none of the students is seating on the seat previously allotted is \(\dfrac{1}{180}\).</p><p>(d) The probability that even numbered students are seating at even numbered seat and none of the students is seating on the seat previously allotted is \(\dfrac{1}{30}\).</p>
<p>The probability that \(S_4\) gets allotted seat \(R_4\) and none of the remaining students gets the seat previously allotted to them, is \(\dfrac{11}{180}\).</p>
<p>The probability that \(S_5\) gets allotted seat \(R_5\) and none of the remaining gets the seat previously allotted to them is \(\dfrac{1}{6}\).</p>
<p>The probability that even numbered students are seating at even numbered seat and none of the students is seating on the seat previously allotted is \(\dfrac{1}{180}\).</p>
<p>The probability that even numbered students are seating at even numbered seat and none of the students is seating on the seat previously allotted is \(\dfrac{1}{30}\).</p>
Step-by-Step Solution
Key Concept: Use derangement formula D(n) = n!∑(−1)^k/k! to count permutations where no element is in its original position. When one student is fixed in their seat, find derangements of remaining students; when positions are restricted (even→even), use conditional probability with derangements.
<p><strong>Derangement Formula:</strong> D(n) = n!∑(−1)^k/k! = n![1 − 1/1! + 1/2! − 1/3! + ...]</p><p>D(3) = 2, D(4) = 9, D(5) = 44</p><p><strong>Part (a): S₄ at R₄, others deranged</strong></p><p>Fix S₄ at R₄. Remaining 5 students must be deranged among remaining 5 seats.</p><p>P = D(5)/6! = 44/720 = 11/180 ✓ <strong>TRUE</strong></p><p><strong>Part (b): S₅ at R₅, others deranged</strong></p><p>Fix S₅ at R₅. Remaining 5 students deranged.</p><p>P = D(5)/6! = 44/720 = 11/180 ≠ 1/6 <strong>FALSE</strong></p><p><strong>Part (c) & (d): Even students at even seats, all deranged</strong></p><p>Even students: S₂, S₄, S₆ must sit at even seats R₂, R₄, R₆ (originally their seats).</p><p>Odd students: S₁, S₃, S₅ must sit at odd seats R₁, R₃, R₅ (originally their seats).</p><p>For a valid arrangement: no student in original seat.</p><p>Even sub-problem: Arrange {S₂, S₄, S₆} in {R₂, R₄, R₆} with derangement = D(3) = 2</p><p>Odd sub-problem: Arrange {S₁, S₃, S₅} in {R₁, R₃, R₅} with derangement = D(3) = 2</p><p>Favorable outcomes = D(3) × D(3) = 2 × 2 = 4</p><p>Total arrangements = 6! = 720</p><p>P = 4/720 = 1/180 ✓ <strong>TRUE (Part c)</strong></p><p>Part (d) with same answer is <strong>FALSE</strong> (duplicate incorrect option)</p><p><strong>∴ Answer: AD</strong></p>
Correct Answer: AD