JEE MAIN - Mathematics (2022 - 29th June Morning Shift - No. 22)
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
18915
Explanation
There are 98 sets of three consecutive integer and 97 sets of four consecutive integers.
Using the principle of inclusion and exclusion,
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
$=97 \times 98+97 \times 98-97=97 \times 195=18915$.
Comments (0)
