Question 1154577
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In mathematical terms, the problem sounds in this way.


We consider all possible permutations of 10 items.

The number of all such permutations is 10! = 3628800.


Then we ask ourselves : how many are there such permutations of 10 items that no one item is at its native place ?


Such permutations (arrangements) are called <U>derangement</U>.

About derangements, see this Wikipedia article

    https://en.wikipedia.org/wiki/Derangement


In this article, you will find some theory about this subject and the formula to calculate the number of derangement of n items.


The formula is quite complicated, but there is a Table in this reference, containing calculated values for moderate values of n.


For n= 10, the number of derangement, from this Table, is equal to 1334961.


So, there are 10! = 3628800 permutations of 10 objects, in all (it is the sample space in this problem).


Of them, there are 1334961 <U>favorable</U> derangement permutations.


The probability under the problems's question is, therefore, 


    P = {{{favorable/total}}} = {{{1334961/3628800}}} = 0.3679  (approximately).     <U>ANSWER</U>
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