JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Permutations and Combinations

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.

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.

The probability that, on the examination day, the student

S_{1}

gets the previously allotted seat

R_{1}

, and NONE of the remaining students gets the seat previously allotted to him/her is

Answer: (A)

Solution

We need

S_{1}

R_{1}

and the remaining 4 students in a derangement of their seats.

Number of derangements of 4 items:

D_{4} = 4! (\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !}) = 24 (\frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 24 \cdot \frac{9}{24} = 9

.

Total arrangements:

5! = 120

.

Probability:

\frac{9}{120} = \frac{3}{40}

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Single correct option