Question 1
In how many ways can 6 distinct letters be placed into 6 envelopes such that exactly 2 letters go into their correct envelopes (and the rest go into wrong envelopes)?
Step-by-Step Solution (Rencontres Numbers):
Concept: This is a partial derangement problem. We want exactly k fixed points, with the remaining n-k elements deranged.
Formula:
$$D(n, k) = \binom{n}{k} \cdot !(n-k)$$
where $!(m)$ is the derangement number.
Given:
- n = 6 (total elements)
- k = 2 (exact fixed points)
Step 1 - Choose which positions are fixed:
Choose 2 positions out of 6: C(6,2) = 15
Step 2 - Derange the remaining elements:
Remaining 4 elements must have NO fixed points
Derangement number D(4) = 9
Step 3 - Apply multiplication principle:
Total = C(6,2) × D(4)
= 15 × 9
= 135
Verification:
- When k=0: D(n,0) = derangement(n) ✓
- When k=n-1: D(n,n-1) = 0 (can't have all but one fixed) ✓
- When k=n: D(n,n) = 1 (identity permutation) ✓
Rencontres numbers table for n=6:
- D(6,0) = 265
- D(6,1) = 264
- D(6,2) = 135
- ... and so on.
Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).
Concept: This is a partial derangement problem. We want exactly k fixed points, with the remaining n-k elements deranged.
Formula:
$$D(n, k) = \binom{n}{k} \cdot !(n-k)$$
where $!(m)$ is the derangement number.
Given:
- n = 6 (total elements)
- k = 2 (exact fixed points)
Step 1 - Choose which positions are fixed:
Choose 2 positions out of 6: C(6,2) = 15
Step 2 - Derange the remaining elements:
Remaining 4 elements must have NO fixed points
Derangement number D(4) = 9
Step 3 - Apply multiplication principle:
Total = C(6,2) × D(4)
= 15 × 9
= 135
Verification:
- When k=0: D(n,0) = derangement(n) ✓
- When k=n-1: D(n,n-1) = 0 (can't have all but one fixed) ✓
- When k=n: D(n,n) = 1 (identity permutation) ✓
Rencontres numbers table for n=6:
- D(6,0) = 265
- D(6,1) = 264
- D(6,2) = 135
- ... and so on.
Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).