There are six students $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$ and $S_{6}$ in music class and for them there are six seats $R_{1}, R_{2}, R_{3}, R_{4}, R_{5}$ and $R_{6}$ arranged in a row, where initially the seat $R_{i}$ is allotted to the student $S_{i}, i = 1, 2, 3, 4, 5, 6$ . But, on the examination day, the six students are randomly allotted the six seats.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

The number of arrangements on the examination day the student $S_{2}$ gets the previously allotted seat $R_{2}$ and none of the remaining students gets the seat previously allotted to him/her is 53.

For i = 1, 2, 3, 4, 5. 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 number of arrangements of the event $T_{1} \cap T_{2} \cap T_{3} \cap T_{4} \cap T_{5}$ is 90.

The number of arrangements on the examination day the student $S_{2}$ gets the previously allotted seat $R_{2}$ and none of the remaining students gets the seat previously allotted to him/her is 44.

answer is A, D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$S_{2}$ gets $R_{2}$ and remaining 5 students does not get the seat allotted to him/her is $D_{5} = 44$
$n (T_{1} \cap T_{2} \cap T_{3} \cap T_{4} \cap T_{5})$ = Total – $n (\bar{T_{1}} \cup \bar{T_{2}} \cup \bar{T_{3}} \cup \bar{T_{4}} \cup \bar{T_{5}})$
= $6! - n (\bar{T_{1}} \cup \bar{T_{2}} \cup \bar{T_{3}} \cup \bar{T_{4}} \cup \bar{T_{5}})$
= $6! - [5 (5!) (2!) - 4 (4!) (2) - 6 (4!) (2!) (2) + 3 (3!) (2) + 5 (3!) (2!) (2!) (2) + 3! (2!) (2!) (2!) - 2 (2!) 2 - 3 (2!) (2!) 2 + 2]$