Disorder problem

Oct 2012
256
19
israel
Hi,
n people hang their coats and hats in a bar.Then ,each one choose randomly a hat and a coat.


What is the probability that:

1)None of them will take his one hat or his own hat?

2)None of them will take both his coat and his hat?

Thank's in advance
 

SlipEternal

MHF Helper
Nov 2010
3,728
1,571
1) look up Derrangement numbers. The answer is $\dfrac{!n}{n!}=\dfrac 1 e$

2) $\dfrac{1}{e^2}$
 

Plato

MHF Helper
Aug 2006
22,462
8,634
Hi,
n people hang their coats and hats in a bar.Then ,each one choose randomly a hat and a coat.
What is the probability that:
1)None of them will take his one hat or his own hat?
2)None of them will take both his coat and his hat?
Please review the statement. It is vague. Are the hat and coat hanged together?
Or are they put on separate racks (the question makes more sense if they are)?
This is a derangement problem.
 
Oct 2012
256
19
israel
can you elaborate your answer,please?
 

Plato

MHF Helper
Aug 2006
22,462
8,634
can you elaborate your answer,please?
@ Hedi, Did you read the references that we provided?
The number of ways $n$ objects can be rearranged so that no one is in its correct position is $\displaystyle\mathscr{D}_n= n!\sum\limits_{k = 0}^n {\dfrac{{{{( - 1)}^k}}}{{k!}}}$
SEE HERE $\mathscr{D}_6=265$ which means if the string $123456$ is randomly rearranged there are $265$ rearrangements in which no one of those six digits is in its correct position.
 
Oct 2012
256
19
israel
The first question is ment to be "nobody will get his own coat or his own hat"
 

Plato

MHF Helper
Aug 2006
22,462
8,634
The first question is meant to be "nobody will get his own coat or his own hat"
You don't answer our sincere questions and moreover you don't think we read English.
I doubt that you can write clear English. So we are left to assume that these men pick a coat at random and then pick a hat at random.
Now because $\sum\limits_{k = 0}^\infty {\dfrac{{{{( - 1)}^k}}}{{k!}}} = \dfrac{1}{e}$
We can use that to model $\mathscr{D}_n=\left\lfloor {\frac{{n!}}{e} + \frac{1}{2}} \right\rfloor \text{ for }n\ge 3$.
Look at this table: SEE HERE

Let $C$ be the event that none of the $n$ men gets his own Coat. Let $H$ be the event that none of the $n$ men gets his own Hat.
$\mathcal{P}(C\cup H)=\mathcal{P}(C)+\mathcal{P}(H)-\mathcal{P}(C\cap H)$ i.e. Does not get his coat or hat.

Now if you were not so self-important you may have done us the honor of explaining the setup of the question.
If there were two different racks, one for coats the other for hats, then clearly the selections would be independent.

So if we assume Independence how do you finish? DON'T ASK!