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Consider a machine whose job is to place 100 letters into 100 envelops.The machine is defective and makes mistakes.What is the probability that in a group of 100 letters no letter is put into the correct envelope?

I did like this.

Total ways of putting letters into envelops=$100!$

And there is only 1 way by which all letter go into correct envelope.

So, I thought my answer must be $\frac{100!-1}{100!}$

But the answer is given to be $\frac{D_{100}}{100!}$

Where $D_{100}$ represents derangement of 100 letters.

Why there is a disparity between my answer and given answer?

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  • $\begingroup$ Consider $3$ letters. The choices are $\color{green}{ABC}, \color{blue}{ACB}, \color{blue}{BAC}, \color{red}{BCA}, \color{red}{CAB}, \color{blue}{CBA}$. $\color{green}{Green} - $ all correct, $\color{blue}{blue} - $ one correct and $\color{red}{red} - $ all incorrect. Note that there is no "two correct" (why?) Your method gives $\frac{3!-1}{3!}=\frac56$, while the given answer gives $\frac{!3}{3!}=\frac26=\frac13$. $\endgroup$ – farruhota Dec 4 '18 at 8:28
  • $\begingroup$ Nitpicking: It appears somewhat bold to conclude that every permutation is equally likely when we are only told that "the machine makes mistakes". (Not to mention that other kinds of mistakes could be that multiple letters end up in the same envelope or some letters are destroyed etc.) $\endgroup$ – Hagen von Eitzen Dec 4 '18 at 18:54
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There are a great many possibilities between "all letters are in the right envelopes" and "no letters are in the right envelopes. For example, we could have five in the right places and the rest in the wrong places. You answered the wrong question there.

As for the derangement problem - I'm sure it's come up plenty of times on the site. See here:
Why is the Derangement Probability so Close to $\frac{1}{e}$?
for a good one, with multiple high quality answers.

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Problem wants no letters in right envelopes. Your approach is to say you want only one in wrong envelope. [Note that can't happen anyway-- if 1 is wrong then 99 are right, but then the last letter has to go in the right envelope.]

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  • $\begingroup$ You mean to say I have calculated atleast 1 letter in wrong envelope? $\endgroup$ – user3767495 Dec 4 '18 at 7:45
  • $\begingroup$ It can't be exactly 1 wrong. What can be said is there is 1 way to get all letters right, and you subtracted that from all ways to stuff envelopes, so looks like you get "at least 1 wrong" which is equivalent really to 2 or more wrong in this problem. $\endgroup$ – coffeemath Dec 4 '18 at 8:11

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