The calculating part is a derangement problem. biffboy's solution is not correct.

]]>biffboy's solution isn't correct.

]]>Why do you calculate that way?

Pls Explain]]>

Welcome to the forum! Please check the link in post #12.

]]>It has been many days since I last posted. Feeling to post again

Anyway,

There had been eight men and eight hats. They were so much drunk that they could not recognise their own hats. So they picked them up in a random manner. Now what is the chance thatno manwill choose his own hat?

Of Course the first step is to find the factorial of 8 which is equal to the number of ways they can pick up the hats in. But it is such a large number that it would take years to write all of them down?Can You give me the answer????

I know the answer but I dont know how to derive it. So a help would be nice

I would have thought the answer was simply 7/8*6/7*5/6*4/5*.3/4*2/3*1/2*1/1=3/4

]]>1. Person i does not take the hat 1. This case is equivalent to solving the problem with n − 1 persons n − 1 hats: each of the remaining n − 1 people has precisely 1 forbidden choice from among the remaining n − 1 hats (i's forbidden choice is hat 1).

2. Person i takes the hat of 1. Now the problem reduces to n − 2 persons and n − 2 hats.

and why hat 1? Why dont we extend the list with Hat 2 and Hat 3 and forever???

]]>Yes.

]]>Do you own a copy of Amusement in Mathematics by Henry. E. Dudeney?]]>

That is a standard book way of getting a recurrence form. I know it as East-West analysis from a famous book written by the late Herbert S. Wilf. Though I have read it many times I am unable to do it or even explain it. I am sorry about that, wish I could do more here.

]]>Suppose that there are n persons numbered 1, 2, ..., n. Let there be n hats also numbered 1, 2, ..., n. We have to find the number of ways in which no one gets the hat having same number as his/her number. Let us assume that first person takes the hat i. There are n − 1 ways for the first person to choose the number i. Now there are 2 options:

1.Person i does not take the hat 1. This case is equivalent to solving the problem with n − 1 persons n − 1 hats: each of the remaining n − 1 people has precisely 1 forbidden choice from among the remaining n − 1 hats (i's forbidden choice is hat 1).2.Person i takes the hat of 1. Now the problem reduces to n − 2 persons and n − 2 hats.From this, the following relation is derived:

!n = (n - 1) (!(n-1) + !(n-2)).\,

Now would you please explain point 1 and 2

]]>That is the recurrence relation for the number of derangements. Did you read this:

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

About half way down the page they will describe the process. It is difficult.

]]>!n=(n-1) (!(n-1) + !(n-2))]]>

It says !n=(n-1)*(!(n-1)+!(n-2))

]]>