There are 5 letters and five addressed envelopes. the number of ways in which…

Concept: A derangement of n distinct objects is a permutation in which NO object lands in its own assigned position. The number of derangements of n objects, written Dn, follows the recurrence Dn = (n - 1)(Dn-1 + Dn-2), starting from D1 = 0 and D2 = 1.

Application: build the derangement count for 5 letters step by step using the recurrence.

1. 1 letter: D1 = 0 (a single letter placed anywhere is always in its own envelope, so no valid derangement exists).

2. 2 letters: D2 = 1 (swapping the two letters between the two envelopes is the only way both go wrong).

3. 3 letters: D3 = (3 - 1)(D2 + D1) = 2 × (1 + 0) = 2.

4. 4 letters: D4 = (4 - 1)(D3 + D2) = 3 × (2 + 1) = 9.

5. 5 letters: D5 = (5 - 1)(D4 + D3) = 4 × (9 + 2) = 4 × 11 = 44.

Cross-check: the closed-form series Dn = n! × sum over k=0..n of (-1)k/k! gives, for n = 5: 120/2! - 120/3! + 120/4! - 120/5! = 60 - 20 + 5 - 1 = 44 exactly, matching the recurrence.

So the number of ways all 5 letters can be placed so that every single one is in the wrong envelope is 44.