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Permutations and Combinations question

2023 · 11 Apr · Shift 1 · Q44
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Permutations and Combinations question

2023 · 11 Apr · Shift 1 · Q44

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
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 ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 44

  1. This is a derangement problem.

    We have 555 students and 555 allotted seats. We want the number of permutations in which no student sits on their own allotted seat.

    So we need the number of derangements of 555 objects, denoted by !5!5!5.

  2. Use the derangement formula:

!n=n!(1−11!+12!−13!+⋯+(−1)n1n!)!n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n\frac{1}{n!}\right)!n=n!(1−1!1​+2!1​−3!1​+⋯+(−1)nn!1​)

For n=5n=5n=5,

!5=5!(1−11!+12!−13!+14!−15!)!5 = 5!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right)!5=5!(1−1!1​+2!1​−3!1​+4!1​−5!1​)
  1. Compute step-by-step:
5!=1205! = 1205!=120

Now evaluate the bracket:

1−1+12−16+124−11201 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}1−1+21​−61​+241​−1201​

Take LCM 120120120:

=0+60−20+5−1120=44120= \frac{0 + 60 - 20 + 5 - 1}{120} = \frac{44}{120}=1200+60−20+5−1​=12044​

Therefore,

!5=120⋅44120=44!5 = 120 \cdot \frac{44}{120} = 44!5=120⋅12044​=44
  1. Hence, the number of ways in which none of the students sits on the allotted seat is
44\boxed{44}44​
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