JEE Advanced 2018 · Paper 1 · Mathematics Q18

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MathematicsSingle correct+3 / -1Question 54 of 54

Passage

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 $T_1 \cap T_2 \cap T_3 \cap T_4$ is

\dfrac{1}{15}

\dfrac{1}{10}

\dfrac{7}{60}

\dfrac{1}{5}

Answer and worked solution

Official answer: C

We need the arrangements of $S_1,\ldots,S_5$ in a row in which no two students with consecutive labels sit next to each other.

Count. Count permutations of $1,2,3,4,5$ with no adjacent consecutive integers by the first element:

Starting with 1, 2, 4 or 5 gives 2, 3, 3 and 2 arrangements.
Starting with 3 gives 4.

For example, 13524, 14253, 24135, 24153, 25314, 31425, 35241, … (and their reversals). The total is 14.

Probability.
$\frac{14}{120}=\frac7{60}.$

(C)

Topics:Classical probability Distribution and arrangements

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