Question

Let b₁, b₂, b₃, b₄ be a 4-element permutation with b_i ∈ {1, 2, 3, .......,100} for 1 ≤ i ≤ 4 and b_i ≠ b_j for i ≠ j, such that either b₁, b₂, b₃ are consecutive integers or b₂, b₃, b₄ are consecutive integers. Then the number of such permutations b₁, b₂, b₃, b₄ is equal to ______.

Options

• 18915

• 18916

• 18917

• 18918

Answer 1

Let b₁, b₂, b₃, b₄ be a 4-element permutation with b_i ∈ {1, 2, 3, .......,100} for 1 ≤ i ≤ 4 and b_i ≠ b_j for i ≠ j, such that either b₁, b₂, b₃ are consecutive integers or b₂, b₃, b₄ are consecutive integers. Then the number of such permutations b₁, b₂, b₃, b₄ is equal to 18915.

Explanation:

b_i ∈ {1, 2, 3, .......,100}

Let P = set when b₁, b₂, b₃ are consecutive.

∴ n(P) = `(97 + 97 + 97 + .... 97)/(98  "times")` = 97 × 98

Let θ = set when b₂, b₃, b₄ are consecutive

∴ n(Q) = `(97 + 97 + 97 + .... 97)/(98  "times")` = 97 × 98

Now, P ∩ Q = Set when b₁, b₂, b₃, b₄ are consecutive.

So, n(P ∪ Q) = n(P) + n(Q) – n(P ∩ Q)

= 97 × 98 + 97 × 98 – 97

= 97(98 + 98 – 1)

= 97(195)

= 18915