TL;DR

A permutation where no element appears in its original position. The number of derangements of n elements is \( D_n = n! \sum_{k=0}^n \frac{(-1)^k}{k!} \approx n!/e \).

By Valenke Exam Prep Team·Last updated 2026-06-03

Derangement

A permutation where no element appears in its original position. The number of derangements of n elements is \( D_n = n! \sum_{k=0}^n \frac{(-1)^k}{k!} \approx n!/e \).

Why it matters for interviews

Classic interview problem: what is the probability a random permutation has no fixed points? The answer \( \approx 1/e \) is elegant and appears in card shuffling, secret Santa, and matching problems.

Definition and Mathematical Foundation

A permutation where no element appears in its original position. The number of derangements of n elements is \( D_n = n! \sum_{k=0}^n \frac{(-1)^k}{k!} \approx n!/e \).

Application in Quantitative Finance

Related Concepts

Derangements

Related Terms

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Frequently Asked Questions

What is the probability that a random permutation is a derangement?▼

\( D_n/n! = \sum_{k=0}^n \frac{(-1)^k}{k!} \to 1/e \approx 0.3679 \). Remarkably, this converges extremely fast -- even for n=5, the probability is within 0.3% of 1/e.

What is the recurrence for derangements?▼

\( D_n = (n-1)(D_{n-1} + D_{n-2}) \) with \( D_0 = 1, D_1 = 0 \). The intuition: element 1 goes to position k (n-1 choices), then either k goes to 1 (leaving D_{n-2}) or not (leaving D_{n-1}).

How is the derangement formula derived?▼

Apply inclusion-exclusion to the events \( A_i \) = 'element i is fixed'. \( |A_{i_1} \cap \cdots \cap A_{i_k}| = (n-k)! \). Summing with alternating signs gives the derangement formula.