Reallocate Seats 2

Required cases = (number of ways of selecting correctly seated student) × (number of ways of de-arranging the others)

The probability is,

$\dfrac {^nC_1 \cdot (n-1)!(\frac {1}{2!}-\frac {1}{3!}+\frac {1}{4!}-\frac {1}{5!}+...(-1)^n\frac {1}{(n-1)!})}{n!}$

$=(\frac {1}{2!}-\frac {1}{3!}+\frac {1}{4!}...$

Using $e^x=1+x+\frac {x^2}{2!}+\frac {x^3}{3!}...$ , we can prove that the above probability(for n tending to ∞) is equal to $e^{-1}$ .

Remove the one student from $n-1$ students whose seat is fixed.

Now, we need to find the probability that none of the rest $n-1$ students are seated in their original positions. This is called Dearrangement.

Derrangement of $p$ people seated in a row such that none of them is alloted their correct positions is $Pp = 1 - \dfrac {1}{1!} + \dfrac {1}{2!} - \dfrac {1}{3!} +....+(-1)^p \dfrac {1}{p! }$

As $p = n-1 \rightarrow \infty -1 \rightarrow \infty,$

$P_\infty = e^{-1}$

As we know Taylor Series of $e^x = 1 + x + \dfrac {x^2}{2!} + \dfrac {x^3}{3!} + \dfrac {x^4}{4!} + .........$

Hence, the required Probabalility $P_n = e^{-1} = \dfrac 1e = \dfrac {1}{2.71428..} = \boxed {0.367...}$