Exercise · ZW-DERANGEMENT-005 · ★★☆ Intermediate
If \(D_n/n!\) approaches \(1/e\), what approximate proportion of permutations of \(n\) elements has at least one fixed point for large \(n\)?
Choose one option.
Show answer
- C. \(\approx 0.63\)
Explanation
Since the derangement proportion tends to \(1/e\), the proportion with at least one fixed point tends to \(1-1/e\approx 0.63\).
Review the concept: Derangement
A derangement is a permutation of \(n\) elements with no fixed points; the number of derangements is \(D_n\).
\[D_n = n!\sum_{k=0}^{n}\frac{(-1)^k}{k!}\]
More practice problems: derangement
- ZW-DERANGEMENT-001 Which of these permutations of \1,2,3,4\, written as σ(1)σ(2)σ(3)σ(4), is a derangement?
- ZW-DERANGEMENT-002 Of the 9 derangements of 4 elements, how many are a single 4-cycle?
- ZW-DERANGEMENT-003 What is D_3?
- ZW-DERANGEMENT-004 What is D_4?
- ZW-DERANGEMENT-006 In the series D_n/n!=∑_k=0^n((-1)^k)/(k!), what is the coefficient of the term k=2?
- ZW-DERANGEMENT-007 For large n, what value does D_n/n! approach?
- ZW-DERANGEMENT-008 How many permutations of 5 elements have exactly one fixed point?
- ZW-DERANGEMENT-009 How many permutations of 4 elements fix at least one point?