Thread: Problem of hats

There are N people. All with their own, unique hat. Each throw his hat into a pile, and mix it. The number of ways for n < N to get their hats is ( N-n)!. Why?

To understand these problems I take N to be a smallish number i.e. N=3, then find the number of arrangements for n=1,2.

Then take N=4 and find the number of arrangements for n=1,2,3

Finally take N=5 and find the number of arrangements for n=1,2,3,4 you'll start to see the pattern.

;D

Originally Posted by phillipshwong

There are N people. All with their own, unique hat. Each throw his hat into a pile, and mix it. The number of ways for n < N to get their hats is ( N-n)!. Why?

Your formula is wrong.

Partial Derangement -- from Wolfram MathWorld

Rencontres numbers - Wikipedia, the free encyclopedia

Hello, phillipshwong!

The problem is badly worded.
Perhaps it was written by an amateur, a student . . .

I think I understand their reasoning . . .


people get their hats back.

The other people may or may not get their hats back.

. . And there are ways to return those hats.

This is very sloppy work!



If they meant exactly get their hats,
. . the problem is more complicated.


The original problem did not consider this.

The other people must not get their own hats.
. . This is derangement of the hats,
. . which can be written:


Example:
. . 8 people, 8 hats.
. . Find the number of ways that exactly 3 get their hats.

There are: . choices of the 3 people.

The other 5 people must not get their own hats.
. . There are: . ways. .**

Therefore, there are: . ways.


**
The Derangement Formula is a topic best left for a separate lesson.
Or you can do a search for "derangement".

Originally Posted by Soroban

Hello, phillipshwong!

The problem is badly worded.
Perhaps it was written by an amateur, a student . . .


I think I understand their reasoning . . .


people get their hats back.

The other people may or may not get their hats back.

. . And there are ways to return those hats.

This is very sloppy work!



If they meant exactly get their hats,
. . the problem is more complicated.


The original problem did not consider this.

The other people must not get their own hats.
. . This is derangement of the hats,
. . which can be written:


Example:
. . 8 people, 8 hats.
. . Find the number of ways that exactly 3 get their hats.

There are: . choices of the 3 people.

The other 5 people must not get their own hats.
. . There are: . ways. .**

Therefore, there are: . ways.


**
The Derangement Formula is a topic best left for a separate lesson.
Or you can do a search for "derangement".

I am waiting for that other lesson.

Originally Posted by phillipshwong

I am waiting for that other lesson.

You need to understand that what Soroban said was very probably a figure of speech and is code for "I don't have the time and MHF is not the place for teaching a course on combinatorics." I suggest you research these things yourself.

That is ok, since I don 't have time for a course, or that concern with d(i). I already figure out why it is ( N-n)!