Let b1b2b3b4 be a 4-element permutation with bi $\in$ {1, 2, 3, ........, 100} for 1 $\le$ i $\le$ 4 and bi $\ne$ bj for i $\ne$ 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 (integer)
18915
Solution
<p>There are 98 sets of three consecutive integer and 97 sets of four consecutive integers.</p>
<p>Using the principle of inclusion and exclusion,</p>
<p>Number of permutations of $b_{1} b_{2} b_{3} b_{4}=$ Number of permutations when $b_{1} b_{2} b_{3}$ are consecutive + Number of permutations when $b_{2} b_{3} b_{4}$ are consecutive - Number of permutations when $b_{1} b_{2}$ $b_{3} b_{4}$ are consecutive</p>
<p>$=97 \times 98+97 \times 98-97=97 \times 195=18915$.</p>
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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