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The problem is as follows, There are 15 passengers, each with their assigned seat. They can take their correct seat, or not.

find probability of A - Atleast one passenger has taken his correct seat.

What I have started with is this: $P(A_i)$ - probability of passenger i taken his correct seat ( while the others can but don't have to) is:

$$P(A_i) = \frac{(n-1)!}{n!} $$

I also think that maybe I should find the opposite, probability of NONE of the passengers taking their correct seat that is $ P(\overline A ) $

However I am stuck here, and don't know how to proceed

wonder
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  • How is it possible that $P(A_i)$ doesn't depend on $i$? – ecrin May 29 '21 at 13:20
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    See Derangements – lulu May 29 '21 at 13:20
  • I guess it's more bad marking, should have said probability of one passenger, not passenger i. As far as derangements, would the answer then be all permutations MINUS number of derangements DIVIDED BY the total number of permutations? – wonder May 29 '21 at 13:34
  • I don;'t understand. There is no $i$ in the problem statement. You introduced it in your attempt. Anyway, yes. Just remove the derangements from the total number of permutations The probability assumption here is that each permutation is equally probable. – lulu May 29 '21 at 13:46

1 Answers1

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It is said that at least one passenger take correct seats, so we can obtain it by all arrangements - all passenger seat wrong places.

We use dearrangement for these types of questions such that Lets assume that one passanger seat correct position and the rest seat randomly , we can do it by $C(15,1) \times 14!$

Now, lets assume that two passanger seat correctly , we can choose correct passangers by $C(15,2) \times 13 !$

Now, lets assume that three passanger seat correctly , we can choose correct passangers by $C(15,3) \times 12 !$

This procces will continue until all passangers are correct , ie, $C(15,15) \times 0! =1$

As you realize when we count the cases , we make overcounting , so we can get rid of it by exclusion - inclusion principle.

So , answer is $15!-C(15,1) \times 14! + C(15,2) \times 13 !-C(15,3) \times 12 !+....-1$

We can shorten this by $n! \times [1-1!+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-..-\frac{1}{n!}]= 15! \times [1-1!+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-..-\frac{1}{15!}]$

Not a Salmon Fish
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  • The expression you have has a shorthand notation which should be mentioned... the final expression you wrote is $n! - !n$ – JMoravitz May 29 '21 at 14:33
  • @JMoravitz i did not understand what you wanted to mean by n!-!n – Not a Salmon Fish May 29 '21 at 15:29
  • $n!$ is the notation for factorials, the number of permutations of an $n$-element set. $!n$ on the other hand is the notation for derangements, the number of permutations of an $n$-element set where the permutations have no fixed points. The expression you have written can be simplified by using this notation for $!n$ being the number of derangements, your long expression $n!\times[1-1!+\frac{1}{2!}-\frac{1}{3}!+\dots\pm\frac{1}{n!}]$ can be written much more concisely as $n!-!n$, read aloud as "n factorial minus n subfactorial." See the wiki link above posted by lulu. – JMoravitz May 29 '21 at 15:31
  • @JMoravitz thankf for this information , i generally use $D_n$ for derangement instead of $!n$ – Not a Salmon Fish May 29 '21 at 15:37