Question 1: Three letters and three addressed envelopes are given. Find the number of arrangements where no letter goes into its correct envelope.

Solution:

Method 1: Listing Arrangements

Total arrangements = \( 3! = 6 \). Let the letters be L1, L2, L3, and envelopes E1, E2, E3, with L1 assigned to E1, etc. The arrangements are:

1. E1-L1, E2-L2, E3-L3 (Correct, not a derangement)
2. E1-L1, E2-L3, E3-L2 (L1 correct, not a derangement)
3. E1-L2, E2-L1, E3-L3 (L3 correct, not a derangement)
4. E1-L2, E2-L3, E3-L1 (Derangement)
5. E1-L3, E2-L1, E3-L2 (Derangement)
6. E1-L3, E2-L2, E3-L1 (L2 correct, not a derangement)

Derangements: 2.

Method 2: Derangement Formula

For \( n = 3 \):

$$!3=3![1–\frac{1}{1!}+\frac{1}{2!}–\frac{1}{3!}]=6·\frac{1}{3}=2$$

Answer: 2 derangements.

Question 2: Six students are appearing for an examination. Find the number of seating arrangements where at least two students are in the wrong seats.

Solution:

Total arrangements = \( 6! = 720 \).

All correct arrangements (no student in the wrong seat) = 1.

Arrangements with at least one student in the wrong seat = \( 720 – 1 = 719 \).

Since swapping two students’ seats results in at least two wrong placements (if student X is in Y’s seat, Y is not in Y’s seat), the condition “at least two students in wrong seats” is equivalent to “at least one student in the wrong seat” in this context, as no single swap can result in exactly one wrong seat. Thus:

Answer: 719 arrangements.

Note: If the question asked for all students in wrong seats, we’d use the derangement formula for \( n = 6 \):

$$!6=6![1–\frac{1}{1!}+\frac{1}{2!}–\frac{1}{3!}+\frac{1}{4!}–\frac{1}{5!}+\frac{1}{6!}]=720·\frac{53}{144}=265$$

Question 3: Find the number of permutations of {1, 2, 3, 4, 5} that keep at least one integer in its original position.

Solution:

Total permutations = \( 5! = 120 \).

Derangements (no integer in its original position) = \( !5 = 44 \).

Permutations with at least one integer fixed = Total permutations – No integer is in correct position (Derangements):

$$120–44=76$$

Answer: 76 permutations.

Question 4: How many permutations of {1, 2, 3, 4, 5} leave at least two elements in their original positions?

Solution:

We calculate the number of permutations where exactly \( k \) elements are fixed (in their original positions) and the remaining \( n – k \) elements are deranged. Sum these for \( k = 2 \) to \( k = 5 \):

For at least 2 elements being fixed, we can consider the following cases:

• Exactly 2 elements are fixed, and the remaining 3 are all in the wrong positions.
• Exactly 3 elements are fixed, and 2 elements are in the wrong positions.
• Exactly 4 elements are fixed, and 1 element is in the wrong position.
• All 5 elements are fixed, and 0 elements are in the wrong position.

• Exactly 2 fixed: Choose 2 elements to be fixed: $C_2^5$ = 10. The remaining 3 elements must be deranged: \( !3 = 2 \). Total = \( 10 \cdot 2 = 20 \).
• Exactly 3 fixed: Choose 3 elements: $C_3^5$ = 10. Remaining 2 elements deranged: \( !2 = 1 \). Total = \( 10 \cdot 1 = 10 \).
• Exactly 4 fixed: Choose 4 elements: $C_4^5$ = 5. Remaining 1 element deranged: \( !1 = 0 \). Total = \( 5 \cdot 0 = 0 \).
• Exactly 5 fixed: Choose 5 elements: $C_5^5$ = 1. No elements deranged: \( !0 = 1 \). Total = \( 1 \cdot 1 = 1 \).

Total = \( 20 + 10 + 0 + 1 = 31 \).

Answer: 31 permutations.

Question 5: Let \( A = \{1, 2, 3, 4, 5\} \). How many injective functions \( f : A \to A \) have the property that for each \( x \in A \), \( f(x) \neq x \)?

Solution:

An injective function \( f : A \to A \) is a permutation of the set \( A \). The condition \( f(x) \neq x \) means no element maps to itself, which is the definition of a derangement. Thus, we need the number of derangements for \( n = 5 \):

$$!5=5![1–\frac{1}{1!}+\frac{1}{2!}–\frac{1}{3!}+\frac{1}{4!}–\frac{1}{5!}]=44$$

Answer: 44 injective functions.

Question 6: Ten ladies drop off their red hats at a museum’s hat check. The attendant returns the hats randomly. In how many ways can exactly six ladies receive their own hats (and the other four not)?

Solution:

Choose 6 ladies who receive their own hats: $C_6^{10}$ = \( \frac{10!}{6! \cdot 4!} = 210 \).

The remaining 4 ladies must not receive their own hats, so we need the number of derangements for the 4 hats: \( !4 = 9 \).

Total ways = \( 210 \cdot 9 = 1890 \).

Answer: 1890 ways.