Let b₁b₂b₃b₄ be a 4-element permutation with bᵢ ∈ 1, 2, 3, ........, 100 for 1 ≤ i ≤ 4 and bᵢ ≠ bⱼ 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 ____________.

Solution

Maths sir · 4 steps

Count each block of three separately, then subtract the cases where both blocks happen together.

1. First three consecutive. Take b₁, b₂, b₃ as k, k+1, k+2. Since k+2 ≤ 100, the start k can be 1, 2, …, 98, which is 98 choices. Then b₄ is any of the 97 numbers not already used, so this gives 98 × 97 = 9506 permutations.
2. Last three consecutive. The same count works for b₂, b₃, b₄ as k, k+1, k+2. There are 98 starts, and b₁ can be any of the 97 unused numbers, so this also gives 9506.
3. Overlap. Both conditions hold only when all four are consecutive: k, k+1, k+2, k+3 with k+3 ≤ 100. That gives 97 permutations, and each one was counted in both groups above, so it must be removed once.

   n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = 9506 + 9506 - 97

4. So. The number of such permutations is 19012 - 97 = 18915.

The idea to keep. Whenever a question says "either this or that", use inclusion–exclusion. Count each case on its own, then remove the overlap that was counted twice.

Answer from the NTA final answer key. Solution by PrepWiser.