SOLUTION: Suppose all 10 men at a party throw their hats in the center of the room. Each man then randomly selects a hat. Find the probability that none of the 10 men selects his own hat.

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 derangement.

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 favorable derangement permutations.


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


    P =  =  = 0.3679  (approximately).     ANSWER