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The question goes like this:

There are 5 envelopes corresponding to 5 letters. If the letters are placed in the envelopes at random, what is the probability that all the letters are not placed in the right envelopes?

The book finds easy way to solve this:

There is only one way to put all the letters in correct envelopes, we can say that event of not all four letters going into the correct envelopes will be given by $5!-1=119$. This the desired probability is $\frac{119}{120}$.

I want to find the other way: counting / summing individual possibilities like 2 letters in wrong envelopes plus 3 letters in wrong plus 4 letters in wrong envelopes plus 5 letters in wrong envelopes. Realizing that n letters can be in wrong envelopes in $\binom{5}{n}\times (n-1)$ ways, I get $\binom{5}{2}\times 1 +\binom{5}{3}\times 2 +\binom{5}{4}\times 3 +\binom{5}{5}\times 4 = 49$. Where I am making mistake?

Mahesha999
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    Let $N$ denote the number of envelopes that are placed in the right envelopes. Then the question seems to be: what is $P(N=0)$? However, the book answers the question: what is $P(N<5)$? – drhab Sep 05 '15 at 17:49
  • It is so much easier to compute $P(N < 5)$ the way the book has done. For the way you are proposing to use, you will have to look up derangements and partial derangements, from, say wolfram. – true blue anil Sep 05 '15 at 17:53
  • @drhab: The wording "probability... all the letters are not placed in the right envelope" means $P(N < 5)$, For $P(N=0)$, the wording would be be none of the letters.. – true blue anil Sep 05 '15 at 17:57
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    @trueblueanil I would expect "not all the letters are placed in the right envelope" for $P(N<5)$ (at least one is not). – drhab Sep 05 '15 at 18:02
  • @drhab: Your wording is certainly the best (and correct) way to express the intent, but I'd interpret the questioner's wording to mean the same. – true blue anil Sep 05 '15 at 18:17
  • hey guys I want to know what I am doing wrong. Does solving it my way calls for using inclusion-exclusion principle? But then how should I apply? @trueblueanil what will be derangements and partial derangements? Solving this way may not be practical for exams but surely will help to make concepts strong. Is it possible to solve this on Wolfram? Please tell. – Mahesha999 Sep 05 '15 at 19:37
  • Yes, you need to use inclusion-exclusion. This is a duplicate to some extent of several previous postings on this site. Use 'letters envelopes' in Search box to see several of them. This is also Problem 1.5 in Suess & Trumbo :Intro to Prob. Simulation...(Springer), you can find it in the the free preview offered on the Google Books site. (Problem is for simulation, note gives answer.) – BruceET Sep 05 '15 at 20:10

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It is not a complete answer. I generated a small program and counted the misses. Let $TM(i)=k$ denote that the number of combinations with $i$ misses is $k$. Then the numbers you are looking for are the following : $$ \eqalign{ TM(2) &= 10 \cr TM(3) &= 20 \cr TM(4) &= 45 = 5 \cdot 3 \cdot 3 \cr TM(5) &= 44 = 4 \cdot 11 \cr } $$ Indeed the total is $119$. (Even though the first two cases were easy to understand, I had problem to conceive the last two).

nickchalkida
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