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JEE 2023 Mathematics PYQ — In an examination, 5 students have been allotted their seats as p… | Mathem Solvex | Mathem Solvex

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

In an examination, 5 students have been allotted their seats as per their roll numbers. The number of ways, in which none of the students sits on the allotted seat is ________.

Choose the correct answer:

A.
44
(Correct Answer)
B.
45
C.
46
D.
54

Correct Answer:

Explanation

Step 1: Formula for Derangements ( $D_n$ )

D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + (-1)^n \frac{1}{n!} \right)

Step 2: Values substitute karein ( $n = 5$ )

D_5 = 5! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right)

(Note: $1 - \frac{1}{1!}$ zero ho jata hai)

Step 3: Calculation

D_5 = 120 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right)

D_5 = \frac{120}{2} - \frac{120}{6} + \frac{120}{24} - \frac{120}{120}

D_5 = 60 - 20 + 5 - 1

D_5 = 44

Explanation

Step 1: Formula for Derangements ( $D_n$ )

D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + (-1)^n \frac{1}{n!} \right)

Step 2: Values substitute karein ( $n = 5$ )

D_5 = 5! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right)

(Note: $1 - \frac{1}{1!}$ zero ho jata hai)

Step 3: Calculation

D_5 = 120 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right)

D_5 = \frac{120}{2} - \frac{120}{6} + \frac{120}{24} - \frac{120}{120}

D_5 = 60 - 20 + 5 - 1

D_5 = 44

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