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There are five students S_(1),S_(2),S_(3),S_(4) and R_(5) arranged in a row, where initially the seat R_(1)is allotted to the students are randomly allotted the five seats R_(1),R_(2),R_(3),R_(4) and R_(6) 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. (There are two questions based on Paragraph, the question given below is one of them) For i=1,2,3,4. let T_(i) denote the event that the students S_(1) 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) cap T_(2) cap T_(3) cap T_(4) is |
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Answer» `(1)/(15)` Total `=-n(bar(T_(1)) CUP bar(T_(2)) cup bar(T_(3)) cup bar(T_(4))) ` `implies n(T_(1) cap T_(2) cap T_(3) cap T_(4))` `=5! -[""^(4)C_(1)4! 2!-(""^(3)C_(1) 3!2!+""^(3)C_(1)3!2!2!)+(""^(2)C_(1)2!2!+""^(4)C_(1) 2*2!)-2]` ` implies n(T_(1) cap T_(2) cap T_(3) cap T_(4))` `=120-[192-(36+72)+(8+16)-2]` `=120-[192-108+24-2]=14` `:.` Required PROBABILITY `=(14)/(120)=(7)/(60)` |
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