There are five students S 1 , S 2 , S 3 , S 4 and S 5 in a music class and for them there are five seats R 1 , R 2 , R 3 , R 4 and R 5 arranged in a row, where initially the seat R i is allotted to the student 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 , let T i denote the event that the students S i and 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 is B A where A , B are co-prime integers. Find A + B
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You wrote down all the possible arrangements but what if number of students is large ?
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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 ) ′
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It is easy, by considering the possible placements for S 1 and S 2 , to find the 1 4 possible arrangements of the five students: 1 4 2 5 3 3 1 4 2 5 2 4 1 3 5 2 5 3 1 4 4 2 5 3 1 1 3 5 2 4 3 1 5 2 4 2 4 1 5 3 4 2 5 1 3 3 5 2 4 1 4 1 3 5 2 3 5 1 4 2 5 2 4 1 3 5 3 1 4 2 and so the desired probability is 5 ! 1 4 = 6 0 7 , making the answer 6 7 .