An algebra problem by utkarsh grover

The number of ways that we can assign $n$ keys to $n$ locks such that no key gets inserted into its appropriate lock is given by the dearrangement formula

$D_n = n! \sum_{r=0}^{n} \dfrac{(-1)^r}{r!}$

Now the total number of permutations possible in our sample space is $S_n = n!$ .

Thus for $n$ locks, the required probability is

$P_n = \dfrac{D_n}{S_n} = \sum_{r=0}^{n} \dfrac{(-1)^r}{r!}$

As $n \to \infty$ , we get

$P_{\infty} = \lim_{n \to \infty} \sum_{r=0}^{n} \dfrac{(-1)^r}{r!} = \sum_{r=0}^{\infty} \dfrac{(-1)^r}{r!} = e^{-1} = \dfrac 1e \approx 0.367879$

The result rounding to nearest tenth gives the answer as $\boxed{0.4}$ .

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Note:

The result $\displaystyle \sum_{r=0}^{\infty} \dfrac{(-1)^r}{r!} = e^{-1}$ follows from the Taylor series for $e^x$ given by

$e^x = \sum_{r=1}^{\infty} \dfrac{x^r}{r!}$

Putting $x=-1$ gives the desired result.