JEE Advanced 2018 - 3

There are five students S 1 S_{1} , S 2 S_{2} , S 3 S_{3} , S 4 S_{4} and S 5 S_{5} in a music class and for them there are five seats R 1 R_{1} , R 2 R_{2} , R 3 R_{3} , R 4 R_{4} and R 5 R_{5} arranged in a row, where initially the seat R i R_{i} is allotted to the student S i , i = 1 , 2 , 3 , 4 , 5 S_{i},i = 1, 2, 3, 4, 5 . But, on the examination day, the five students are randomly allotted the five seats.

For i = 1 , 2 , 3 , 4 i = 1, 2, 3, 4 , let T i T_{i} denote the event that the students S i S_{i} and S i + 1 S_{i+1} do NOT sit adjacent to each other on the day of the examination. Then the probability of the event T 1 T 2 T 3 T 4 T_{1}\cap T_{2}\cap T_{3}\cap T_{4} is A B \frac{A}{B} where A , B A,B are co-prime integers. Find A + B A+B


The answer is 67.

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1 solution

Mark Hennings
May 22, 2018

It is easy, by considering the possible placements for S 1 S_1 and S 2 S_2 , to find the 14 14 possible arrangements of the five students: 14253 13524 31425 31524 41352 24135 24153 35142 53142 25314 42513 52413 42531 35241 \begin{array}{cccc} 14253 & 13524 \\ 31425 & 31524 & 41352 \\ 24135 & 24153 & 35142 & 53142 \\ 25314 & 42513 & 52413 \\ 42531 & 35241 \end{array} and so the desired probability is 14 5 ! = 7 60 \tfrac{14}{5!} = \tfrac{7}{60} , making the answer 67 \boxed{67} .

You wrote down all the possible arrangements but what if number of students is large ?

Shivam Jadhav - 3 years ago

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Then I would use the Inclusion-Exclusion Principle to count the number of permutations in T 1 T 2 T 3 T n 1 = ( T 1 T 2 T n 1 ) T_1' \cup T_2' \cup T_3' \cup \cdots \cup T_{n-1}' = (T_1 \cap T_2 \cap \cdots T_{n-1})'

Mark Hennings - 3 years ago

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Same question with 8 students

diwakar kumar - 2 years, 11 months ago

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