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 _______ .
Correct Answer: 18915
Solution and Explanation
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.
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Concepts Used:
Permutations and Combinations
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