Permutations & Combinations
Derangements
Grade 11
Question:
<p>Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is</p>
<p>264</p>
<p>265</p>
<p>53</p>
<p>67</p>
Step-by-Step Solution
Key Concept: Fix card 1 in envelope 2, then find derangements of the remaining 5 cards in 5 envelopes where card i cannot go in envelope i. This reduces to finding D₅ (derangements of 5 objects).
<p><strong>Step 1:</strong> Fix card 1 in envelope 2 (given condition). Now we have 5 remaining cards {2,3,4,5,6} and 5 remaining envelopes {1,3,4,5,6}.</p><p><strong>Step 2:</strong> For envelope 1: card 1 cannot go there (already placed). Card 2 cannot go in envelope 2 (since envelope 2 is taken by card 1). So we need a derangement where card i ≠ envelope i for i ∈ {2,3,4,5,6}.</p><p><strong>Step 3:</strong> Calculate D₅ (derangements of 5 objects):<br/>D₅ = 5!∑(−1)ᵏ/k! = 120(1 − 1 + 1/2 − 1/6 + 1/24 − 1/120)<br/>D₅ = 120(44/120) = 44</p><p><strong>Step 4:</strong> Verify using recurrence: D₅ = 4(D₄ + D₃) = 4(9 + 2) = 4(11) = 44</p><p>∴ Answer: <strong>44</strong></p>
Correct Answer: C