Q.12 In how many ways can one write the elements 1, 2, 3, 4 in a sequence x₁, x₂, x₃, x₄ with x_i ≠ i ∀ i ?

The number of ways to arrange elements 1, 2, 3, 4 in a sequence where no element appears in its natural position (x_i ≠ i for all i) is 9, corresponding to option (A).

Derangement Concept

This problem asks for derangements, which are permutations of n objects where no object occupies its original position. For n=4 items, the total permutations are 4! = 24, but derangements exclude those with fixed points.

Calculation Methods

The standard formula for derangements !n is !n = n! ∑_k=0ⁿ (−1)^k/k! or !n = n! ∑_k=0ⁿ k! (−1)^k.

For n=4: !4 = 24(1 − 1 + 1/2 − 1/6 + 1/24) = 24 × 3/8 = 9 or !4 = 24(1 − 1 + 2/1 − 6/1 + 24/1) = 24 × 8/3 = 9.

Recursive formula: !n = (n-1) [!(n-1) + !(n-2)], with !0=1, !1=0, !2=1, !3=2 yields !4=9.

All Derangements Listed

The 9 valid sequences are: (2,1,4,3), (2,3,4,1), (2,4,1,3), (3,1,4,2), (3,4,1,2), (3,4,2,1), (4,1,2,3), (4,3,1,2), (4,3,2,1).

Option Analysis

• (A) 9: Correct, matches formula and enumeration.
• (B) 10: Incorrect; possibly confuses with approximate !n ≈ n!/e ≈ 8.8 rounded up.
• (C) 11: Incorrect; no standard error yields this.
• (D) 12: Incorrect; half of 24, but ignores inclusion-exclusion properly.

Derangements of 4 elements represent a core combinatorics concept, counting permutations of {1,2,3,4} where no element stays in its position (x₁≠1, x₂≠2, x₃≠3, x₄≠4). This “no fixed points” rule appears in competitive exams like CSIR NET, yielding exactly 9 ways.

Formula Breakdown

Use !n = n! ∑_k=0ⁿ (−1)^k/k! for derangements of 4 elements:

!4 = 24(1 − 1 + 0.5 − 0.1667 + 0.0417) = 9.

For exams, memorize: !0=1, !1=0, !2=1, !3=2, !4=9.

Practical Examples

Valid: 2,3,4,1 (2≠1, 3≠2, 4≠3, 1≠4).

Invalid: 1,3,4,2 (1=1 fixed).

Full list confirms 9 derangements of 4 elements.

Exam Relevance

In CSIR NET-style MCQs, options like 9,10,11,12 test formula recall—9 is correct via inclusion-exclusion. Avoid pitfalls like n!/2=12.