Question

Let b₁b₂b₃b₄ 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 b₁b₂b₃b₄ is equal to _______ .

Answer 1

Correct Answer: 18915

98 sets of three consecutive integer and 97 sets of four consecutive integers.
By Using the principle of inclusion and exclusion,
The number of permutations of b₁b₂b₃b₄ = The number of permutations when b₁b₂b₃ are consecutive + The number of permutations when b₂b₃b₄ are consecutive – The number of permutations when b₁b₂b₃b₄ are consecutive.
=97 × 98 + 97 × 98 – 97 = 97 × 195
= 18915

So, the answer is 18915.