1. Tardigrade
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  3. Mathematics
  4. PARAGRAPH A There are five students S1, S2, S3, S4 and S5 in a music class and for them there are five seats R1, R2,R3,R4 and R5 arranged in a row, where initially the seat Ri is allotted to the student Si, i=1,2,3,4,5. But, on the examination day, the five students are randomly allotted the five seats. Question:For i=1,2,3,4, let Ti denote the event that the students Si and Si+1 do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event T1 ∩ T2 ∩ T3 ∩ T4 is

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Q. PARAGRAPH "A"
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.
Question: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 $T_1 \cap T_2 \cap T_3 \cap T_4$ is

JEE AdvancedJEE Advanced 2018

Solution:

$n\left(T_{1} \cap T_{2} \cap T_{3} \cap T_{4}\right) = Total - n \left(\bar{T_{1}} \cup \bar{T_{2} }\cup \bar{T_{3}} \cup \bar{T_{4}}\right) $
$= 5! - \left(^{4} C_{1} 4! 2! - \left(^{3}C_{1} . 3!2!+^{3}C_{1}3!2!2!\right) + \left(^{2}C_{1} 2!2! + ^{4}C_{1} .2.2!\right)-2\right) $
$ = 14$
Probability $ = \frac{14}{5!} = \frac{7}{60}$