41. In how many ways can one write the elements 1, 2, 3, 4 in a sequence x1, x2, x3, x4 with xi G i 6i?

(A) 9

(B) 10

(C) 11

(D) 12

Number of Ways to Arrange 1, 2, 3, 4 Such That xᵢ ≠ i for Every i

Understanding the Given Arrangement Problem

This question asks us to arrange the four elements 1, 2, 3, and 4 in a sequence:

x₁, x₂, x₃, x₄

The condition:

x_i ≠ i for all i

means that no element can appear in its original position. Therefore, 1 cannot occupy the first position, 2 cannot occupy the second position, 3 cannot occupy the third position, and 4 cannot occupy the fourth position.

An arrangement in which none of the elements occupies its original position is called a derangement. Therefore, this problem is asking for the number of derangements of four distinct elements.

What Is a Derangement?

A derangement is a permutation in which no element remains in its original position. Suppose the original order of the elements is:

1, 2, 3, 4

In a valid arrangement, the first position cannot contain 1, the second position cannot contain 2, the third position cannot contain 3, and the fourth position cannot contain 4.

For example, the arrangement:

(2, 1, 4, 3)

is valid because:

x₁ = 2 ≠ 1

x₂ = 1 ≠ 2

x₃ = 4 ≠ 3

x₄ = 3 ≠ 4

Therefore, no element is present in its original position.

However, the arrangement:

(2, 1, 3, 4)

is not valid because 3 is in the third position and 4 is in the fourth position. Thus, it violates the condition x_i ≠ i.

Finding the Total Number of Permutations

Before applying the restriction, the four distinct elements 1, 2, 3, and 4 can be arranged in:

4! = 4 × 3 × 2 × 1

Therefore:

4! = 24

However, many of these 24 permutations contain one or more elements in their original positions. We need to count only those permutations in which no element occupies its corresponding position.

Using the Derangement Formula

The number of derangements of n distinct elements is denoted by !n or D_n. The standard derangement formula is:

D_n = n![1 − 1/1! + 1/2! − 1/3! + … + (−1)ⁿ/n!]

For four elements:

D₄ = 4![1 − 1/1! + 1/2! − 1/3! + 1/4!]

Since:

4! = 24

we get:

D₄ = 24[1 − 1 + 1/2 − 1/6 + 1/24]

Now:

1 − 1 = 0

Therefore:

D₄ = 24[1/2 − 1/6 + 1/24]

Taking 24 as the common denominator inside the brackets:

1/2 = 12/24

1/6 = 4/24

Hence:

D₄ = 24[(12 − 4 + 1)/24]

Therefore:

D₄ = 24 × 9/24

Thus:

D₄ = 9

Solving the Problem Using the Inclusion-Exclusion Principle

The result can also be derived in detail using the inclusion-exclusion principle. This method begins with all possible permutations and systematically removes arrangements containing fixed points.

Let:

A₁ = arrangements in which 1 is in position 1

A₂ = arrangements in which 2 is in position 2

A₃ = arrangements in which 3 is in position 3

A₄ = arrangements in which 4 is in position 4

We need arrangements that do not belong to any of these four sets.

Counting Arrangements With at Least One Specified Fixed Element

If one particular element is fixed in its original position, the remaining three elements can be arranged in:

3! = 6

ways. Since any one of the four elements may be selected as the fixed element, the total contribution is:

C(4, 1) × 3! = 4 × 6 = 24

Counting Arrangements With Two Specified Fixed Elements

If two particular elements are fixed, the remaining two elements can be arranged in:

2! = 2

ways. The two fixed elements can be selected in:

C(4, 2) = 6

ways. Therefore, the total contribution is:

C(4, 2) × 2! = 6 × 2 = 12

Counting Arrangements With Three Specified Fixed Elements

If three elements are fixed in their original positions, the remaining element is automatically fixed as well. The three elements can be selected in:

C(4, 3) = 4

ways, and the remaining one element can be arranged in:

1! = 1

way. Therefore, the contribution is:

C(4, 3) × 1! = 4

Counting the Arrangement With All Four Elements Fixed

If all four elements are fixed, there is only one arrangement:

(1, 2, 3, 4)

Therefore, the contribution is:

C(4, 4) × 0! = 1

Applying Inclusion-Exclusion

According to the inclusion-exclusion principle, the number of permutations with no fixed points is:

D₄ = 4! − C(4,1)3! + C(4,2)2! − C(4,3)1! + C(4,4)0!

Substituting the values:

D₄ = 24 − 24 + 12 − 4 + 1

Therefore:

D₄ = 9

Direct Verification by Listing All Valid Arrangements

Since there are only four elements, the answer can also be verified by listing every valid arrangement in which no element appears in its original position.

The nine valid arrangements 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)

Each of these arrangements satisfies:

x₁ ≠ 1, x₂ ≠ 2, x₃ ≠ 3, and x₄ ≠ 4

Therefore, the direct counting method also confirms that the total number of valid sequences is 9.

Analysis of All the Given Options

Option (A): 9

This option is correct. The required arrangements are derangements of four elements. Using either the derangement formula, the inclusion-exclusion principle, or direct listing, the total number of valid sequences is 9.

Option (B): 10

This option is incorrect. There are only nine permutations in which all four elements are displaced from their original positions. A count of 10 would include at least one arrangement containing a fixed element.

Option (C): 11

This option is incorrect. The inclusion-exclusion calculation gives 24 − 24 + 12 − 4 + 1 = 9, not 11.

Option (D): 12

This option is incorrect. Although the term C(4,2) × 2! = 12 appears during the inclusion-exclusion calculation, it represents arrangements associated with two specified fixed positions and is not the final number of derangements.

Final Answer

The number of ways to arrange the elements 1, 2, 3, and 4 such that:

x_i ≠ i for all i

is:

9

Correct Option: (A) 9