Question

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 _______ .

Answer 1

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.