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There are 6 letters and 6 houses. Find Probability that

  • no house receives the correct letter.

  • at least one house gets a correct letter.

1 )A=No house receives the correct letter......this is derangements

So, I counted favorable outcomes by derangement formula just like below.

$$D(6)=6!\left(\frac1{0!}-\frac1{1!}+\frac1{2!}-\frac1{3!}+\frac1{4!}-\frac1{5!}+\frac1{6!}\right)=265$$

$P(A)=265/720$

2)B=At least one house gets a correct letter.

$P(B)=1-P(A)=1-265/720=455/720$

Here I want to know whether I am correct in the second scene or not ?? If not how to approach this problem??

Any suggestions will be welcome.

  • 1
    Your approach to (2) is correct and efficient. An alternative to $720-265=455$ is to sum [rencontres numbers](https://en.wikipedia.org/wiki/Rencontres_numbers) $264+135+40+15+0+1=455$ but calculating these involves more work – 2017-01-09
  • 1
    Assuming only one letter is considered "correct" for each house, that no letter can be given out twice, that the distribution of letters is uniformly random, and that each letter is correct for exactly one house, both your answers are correct. – 2017-01-09
  • 0
    Thank you for justification!!!!! @Henry – 2017-01-09
  • 0
    Is there any good resource to comprehend Rencontres numbers apart from Wikipedia @Henry – 2017-01-09
  • 0
    @jerry [OEIS A008290](https://oeis.org/A008290) has several references – 2017-01-09

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