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Partial Derangement (Rencontres Numbers)

Partial Derangement (Rencontres Numbers) counts permutations of n elements with exactly k fixed points (elements that stay in their original position). The formula is D(n,k) = C(n,k) × !(n-k), where !(n-k) is the derangement number. These generalize derangements (k=0) to allow some fixed points.

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What You'll Learn

Introduction & Overview Why This Matters How to Solve Pro Tips & Tricks
Shortcut Methods Common Mistakes Practice Worksheets Exam Importance

Introduction to Partial Derangement (Rencontres Numbers)

Partial Derangement (Rencontres Numbers) counts permutations of n elements with exactly k fixed points (elements that stay in their original position). The formula is D(n,k) = C(n,k) × !(n-k), where !(n-k) is the derangement number. These generalize derangements (k=0) to allow some fixed points.

Prerequisites

Derangement concept Combination formula Derangement numbers Rencontres numbers
Why This Matters: Partial derangement problems appear in 0-1 questions in advanced exams like CAT. They test extension of derangement concepts.

How to Solve Partial Derangement (Rencontres Numbers) Problems

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Step 1: Identify total elements (n) and required fixed points (k)

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Step 2: Choose which k positions will be fixed: C(n, k)

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Step 3: The remaining (n-k) elements must be deranged (no fixed points)

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Step 4: Use derangement number !(n-k) for the remaining elements

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Step 5: Multiply: D(n,k) = C(n,k) × !(n-k)

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Step 6: For k = n, D(n,n) = 1 (identity permutation)

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Step 7: Verify that sum of D(n,k) for k=0 to n equals n!

Pro Strategy: First choose which positions are fixed, then derange the rest. The derangement number for the remaining count gives the number of ways to arrange them with no fixed points. This method works for any k from 0 to n.

Example Problem

Example: How many permutations of 4 elements have exactly 1 fixed point? Solution: Step 1: n = 4, k = 1 Step 2: Choose fixed point: C(4,1) = 4 Step 3: Derange remaining 3 elements: !3 = 2 Step 4: Total = 4 × 2 = 8 Answer: 8 permutations

Pro Tips & Tricks

  • D(n,k) = C(n,k) × !(n-k)
  • Sum of D(n,k) for k=0 to n = n!
  • For k = n-1: D(n,n-1) = 0 (impossible to have n-1 fixed points)
  • For k = n: D(n,n) = 1 (the identity permutation)
  • D(n,0) = !n (derangement)
  • Rencontres numbers form a triangular array

Shortcut Methods to Solve Faster

D(n,1) = n × !(n-1)
D(n,2) = C(n,2) × !(n-2)
D(4,1) = 4 × 2 = 8, D(4,2) = 6 × 1 = 6, D(4,3) = 0, D(4,4) = 1
D(5,1) = 5 × 9 = 45, D(5,2) = 10 × 2 = 20, D(5,3) = 10 × 0 = 0

Common Mistakes to Avoid

Using derangement formula for all elements when fixed points are allowed
Forgetting to multiply by C(n,k) when choosing fixed positions
Confusing rencontres numbers with derangements
Not realizing that D(n,n-1) = 0

Exam Importance

Partial Derangement (Rencontres Numbers) is an important topic for various competitive exams. Here's how frequently it appears:

SSC CGL
0-1 questions
BANKING PO
0-1 questions
RAILWAYS RRB
0-1 questions
CAT
1-2 questions
INSURANCE
0-1 questions

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20 practice questions
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