Partial Derangement (Rencontres Numbers) | Worksheet 3/10

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).

Question 2

In how many ways can 5 distinct letters be placed into 5 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 = 5 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 3

In how many ways can 7 distinct letters be placed into 7 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 = 7 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 7: C(7,2) = 21

Step 2 - Derange the remaining elements:
Remaining 5 elements must have NO fixed points
Derangement number D(5) = 44

Step 3 - Apply multiplication principle:
Total = C(7,2) × D(5)
= 21 × 44
= 924

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=7:
- D(7,0) = 1854
- D(7,1) = 1855
- D(7,2) = 924
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 4

In how many ways can 5 distinct letters be placed into 5 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 = 5 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 5

In how many ways can 7 distinct letters be placed into 7 envelopes such that exactly 1 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 = 7 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 7: C(7,1) = 7

Step 2 - Derange the remaining elements:
Remaining 6 elements must have NO fixed points
Derangement number D(6) = 265

Step 3 - Apply multiplication principle:
Total = C(7,1) × D(6)
= 7 × 265
= 1855

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=7:
- D(7,0) = 1854
- D(7,1) = 1855
- D(7,2) = 924
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 6

In how many ways can 5 distinct letters be placed into 5 envelopes such that exactly 1 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 = 5 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 5: C(5,1) = 5

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(5,1) × D(4)
= 5 × 9
= 45

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 7

In how many ways can 5 distinct letters be placed into 5 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 = 5 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 8

In how many ways can 6 distinct letters be placed into 6 envelopes such that exactly 1 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 = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 6: C(6,1) = 6

Step 2 - Derange the remaining elements:
Remaining 5 elements must have NO fixed points
Derangement number D(5) = 44

Step 3 - Apply multiplication principle:
Total = C(6,1) × D(5)
= 6 × 44
= 264

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).

Question 9

In how many ways can 5 distinct letters be placed into 5 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 = 5 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 10

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).

Question 11

In how many ways can 7 distinct letters be placed into 7 envelopes such that exactly 1 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 = 7 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 7: C(7,1) = 7

Step 2 - Derange the remaining elements:
Remaining 6 elements must have NO fixed points
Derangement number D(6) = 265

Step 3 - Apply multiplication principle:
Total = C(7,1) × D(6)
= 7 × 265
= 1855

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=7:
- D(7,0) = 1854
- D(7,1) = 1855
- D(7,2) = 924
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 12

In how many ways can 7 distinct letters be placed into 7 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 = 7 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 7: C(7,2) = 21

Step 2 - Derange the remaining elements:
Remaining 5 elements must have NO fixed points
Derangement number D(5) = 44

Step 3 - Apply multiplication principle:
Total = C(7,2) × D(5)
= 21 × 44
= 924

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=7:
- D(7,0) = 1854
- D(7,1) = 1855
- D(7,2) = 924
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 13

In how many ways can 5 distinct letters be placed into 5 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 = 5 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 14

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).

Question 15

In how many ways can 7 distinct letters be placed into 7 envelopes such that exactly 1 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 = 7 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 7: C(7,1) = 7

Step 2 - Derange the remaining elements:
Remaining 6 elements must have NO fixed points
Derangement number D(6) = 265

Step 3 - Apply multiplication principle:
Total = C(7,1) × D(6)
= 7 × 265
= 1855

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=7:
- D(7,0) = 1854
- D(7,1) = 1855
- D(7,2) = 924
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 16

In how many ways can 6 distinct letters be placed into 6 envelopes such that exactly 1 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 = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 6: C(6,1) = 6

Step 2 - Derange the remaining elements:
Remaining 5 elements must have NO fixed points
Derangement number D(5) = 44

Step 3 - Apply multiplication principle:
Total = C(6,1) × D(5)
= 6 × 44
= 264

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).

Question 17

In how many ways can 5 distinct letters be placed into 5 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 = 5 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 18

In how many ways can 5 distinct letters be placed into 5 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 = 5 (total elements)
- k = 2 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 2 positions out of 5: C(5,2) = 10

Step 2 - Derange the remaining elements:
Remaining 3 elements must have NO fixed points
Derangement number D(3) = 2

Step 3 - Apply multiplication principle:
Total = C(5,2) × D(3)
= 10 × 2
= 20

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 19

In how many ways can 5 distinct letters be placed into 5 envelopes such that exactly 1 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 = 5 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 5: C(5,1) = 5

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(5,1) × D(4)
= 5 × 9
= 45

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Question 20

In how many ways can 5 distinct letters be placed into 5 envelopes such that exactly 1 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 = 5 (total elements)
- k = 1 (exact fixed points)

Step 1 - Choose which positions are fixed:
Choose 1 positions out of 5: C(5,1) = 5

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(5,1) × D(4)
= 5 × 9
= 45

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=5:
- D(5,0) = 44
- D(5,1) = 45
- D(5,2) = 20
- ... and so on.

Key Insight: The sum of all rencontres numbers for given n equals n! (total permutations).

Master Partial Derangement (Rencontres Numbers) - Beginner Level Problems Partial Derangement (Rencontres Numbers) BEGINNER

What you'll learn in this worksheet:

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20