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?
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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
Your formula is wrong.Quote:
Originally Posted by phillipshwong
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 . . .
Quote:
$\displaystyle \text{There are }N\text{ people, each with a unique hat.}$
$\displaystyle \text{The hats are mixed and returned to them at random.}$
$\displaystyle \text{The number of ways for }n\text{ people to get their hats is }(N-n)!\;\text{ Why?}$
I think I understand their reasoning . . .
$\displaystyle \,n$ people get their hats back.
The other $\displaystyle N-n$ people may or may not get their hats back.
. . And there are $\displaystyle (N-n)!$ ways to return those hats.
This is very sloppy work!
If they meant exactly $\displaystyle \,n$ get their hats,
. . the problem is more complicated.
$\displaystyle \text{There are: }\:_NC_n \:=\:{N\choose n}\text{ ways to select the }n \text{ lucky people.}$
The original problem did not consider this.
The other $\displaystyle N-n$ people must not get their own hats.
. . This is derangement of the $\displaystyle N-n$ hats,
. . which can be written: $\displaystyle d(N-n).$
Example:
. . 8 people, 8 hats.
. . Find the number of ways that exactly 3 get their hats.
There are: .$\displaystyle _8C_3 \:=\:{8\choose3} \:=\:56$ choices of the 3 people.
The other 5 people must not get their own hats.
. . There are: .$\displaystyle d(5) \:=\:44$ ways. .**
Therefore, there are: .$\displaystyle 56\cdot44 \:=\:2464$ 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.Quote:
Originally Posted by Soroban
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.Quote:
Originally Posted by phillipshwong
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)!
