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Question #40585

Difficulty: 8/10 | 🟢 Free

The Question

Let $D_k$ denote the number of derangements of $k$ distinct objects, with $D_0 = 1$ . Define the sequence $S_n$ as: $S_n = \sum_{k=0}^{n} \binom{n}{k} D_k (-1)^{n-k}$ Which of the following represents the value of $\lim_{n \to \infty} \frac{S_n}{n!}$ ?

Options

A.
e^{-2}
B.
e^{-1}
C.
e^{2}
D.
0

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Derangements (Inclusion-Exclusion)|Properties of Binomial Coefficients|General Term Formulation

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of combinatorics (derangements) and analytical series.

The primary trap lies in the 'General Term Formulation'.

Students often attempt to find a recurrence relation for $S_n$ or misapply the inclusion-exclusion principle directly, leading to algebraic dead ends.

The correct path requires expanding $D_k$ and re-indexing the resulting double summation as a Cauchy product.

A critical failure point is not recognizing $\frac{1}{i!(m-i)!}$ as a disguised binomial coefficient $\frac{1}{m!}\binom{m}{i}$ , which is necessary to collapse the inner sum into $2^m$ .

This mimics the process of extracting coefficients from the product of exponential generating functions ( $e^{-x}$ and $\frac{e^{-x}}{1-x}$ ), a classic JEE Advanced maneuver.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Derangements (Inclusion-Exclusion)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free