Fun of Derangement 2

Probability Level 3

How many ways can we arrange the eight digits 1, 2, 3, 4, 5, 6, 7, and 8 so that none of the digits is its position?

For example:

  • 87654321 is acceptable.
  • 18765432 is not acceptable as 1 and 5 are in their respective positions.


The answer is 14833.

Tushar Showrav by Tushar Showrav , 22, Bangladesh

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1 solution

*Relevant wiki:" Derangements .

In this case, the number of ways (using this wiki) is:

1.- the nearest integer to 8 ! e = \frac{8!}{e} = = ⌊ 8 ! e ⌉ = 14833 =\lfloor \frac{8!}{e} \rceil = 14833 2.- 8 ! ( ∑ r = 0 8 ( − 1 ) r r ! ) = 14833 8! \left(\displaystyle\sum_{r = 0}^8 \frac{(-1)^r}{r!}\right) = 14833

3.- We can also use D ( n ) = n D ( n − 1 ) + ( − 1 ) n D(n) = nD(n -1) + (-1)^{n} ... D ( 1 ) = 0 , D ( 2 ) = 1 , . . . D(1) = 0, D(2) =1,...

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