A postman has to deliver 4 letters to a 4 places, he delivers all of them to the wrong places. What is the number of cases in which it is true?

- Ehsan18

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- katieb

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- anonymous

suppose A,B,C,D represent the letters mailed in the correct order to the 4 people.
How many ways can we send letters so that each one receives the wrong letter
B,C,D,A
C,D,A,B
D,A,B,C
B,A,D,C
D,C,B,A...

- wolf1728

In which what is true? You mean in how many ways can he deliver 4 letters to the wrong 4 places?
This has to do with a math concept known as "derangements".
We could figure this by trial and error.
We have four letters which ALL must get delivered to the incorrect address.
So let's say letter 1 cannot go to address 1, letter 2 cannot go to address 2 and so on.
Basically, we have to determine how many ways we can arrange the numbers 1 2 3 and 4
WITHOUT 1 in place 1, 2 is NOT in place 2, etc.
There are NINE such combinations:
2143
2341
2413
3142
3412
3421
4123
4312
4321
Is there a formula? Yes.
Number of derangements = n! * (1-1/1! +1/2! -1/3! + 1/4!)
Number of derangements = 4! * (1 -1 + (1/2) -(1/6) + (1/24)
Number of derangements = 24 * (0 +(1/2) -(1/6) + (1/24)
Lowest common denominator of the fractions = 24
Number of derangements = 24 * (0 +(12/24) -(4/24) + (1/24)
Number of derangements = 24 * (9/24)
Number of derangements = 9
For a calculator and more explanation go here:
http://www.1728.org/derange.htm

- wolf1728

LOL jayzdd gets the medal for an INCORRECT answer ??? !!!
The answer is NINE!!
LOL

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- wolf1728

Heck, I'LL send a medal to jayzdd too!!!

- anonymous

I did not finish

- wolf1728

Well that is okay. I was not really assigning any blame to you - just those medal awarders.
(And why are medals awarded BEFORE someone finishes?)

- anonymous

Using numbers, assuming 1,2,3,4 is the correct order.
I get the same
2,1,4,3
2,3,4,1
2,4,1,3
3,1,4,2
3,4,1,2
3,4,2,1
4,1,2,3
4,3,1,2
4,3,2,1

- anonymous

Nice derangement formula. I did this by making a tree which is tedious

- wolf1728

Hello to ganeshie8 :-)

- wolf1728

Or you could have just copied my numbers :-)

- anonymous

|dw:1434277905759:dw|

- anonymous

some of these branches won't work, because we are left with a 4 in the last branch

- wolf1728

and anyway, to continue with your ABCD method the ones remaining are:
BDAC
CADB
CDBA
DCAB

- anonymous

|dw:1434278137565:dw|

- anonymous

I like the tree way because you can use your fingers to cover the possibilities already covered , systematically.

- wolf1728

Okay Jayzdd
you stuck it out with the "tree" method.
I admire your persistence. That is a good trait.

- anonymous

Thankyou . we came out with the same exact solution. Maybe your trial and error was similar?

- wolf1728

Yes. And actually, I had already done that on my web page.
By the way is OpenStudy acting kind of wacky for you at the moment?
By that I mean, it seems to be "lagging".
When I type a word it fills in the last few letters in a couple of seconds.

- wolf1728

And scrolling down, the scroll bar stops about half way down.

- anonymous

Sometimes it gets laggy because of cache or scripts running in background . You might have to restart your browser, thats what I usually.

- anonymous

do*

- anonymous

Do you know how they derived that derangements formula?

- wolf1728

Hmm okay. (Here I am trying to blame on everything else).
Seriously, it seems you have done well. (And you are up to 58 points in your rating).
This is a rather nice place to "hang out". Yeah Facebook might be okay but here you can actually LEARN something.

- anonymous

Thats your website . Nice

- wolf1728

Shucks thanks jayzdd
I don't know how they got that formula. When I was writing that page (a few years ago) I found the formula probably on Wikipedia or some similar page.

- anonymous

oh :)

- wolf1728

Heck I bet you never knew about "derangements" before tonight right?

- anonymous

I was thinking of approaching this with a complement method.
# of ways each is in wrong place
= # of ways all in the right place - ( # ways all except 1 in the right place + ... )
= n! - (

- anonymous

I'll research this.

- wolf1728

Heck, it's okay.
I was searching on another of my web pages to see if I had an easier derangement formula.
I know I found one (no proof - but it works)
Hold on about a minute - I'll see if I can find it.

- Ehsan18

its nine I get it.....Thanks for the answer

- wolf1728

Hey I found it
Go here:
http://1728.org/puzzle2.htm
and scroll down to puzzle 26
It looks involved and I don't explain the formula that much but it is MUCH easier than the formula I showed at the top of this page.

- wolf1728

And yes ehsan18
it is nine derangements and you'll probably see about a dozen explanations why it is nine LOL

- anonymous

If i follow that , that will explain the process?

- wolf1728

Yes, I'll go to it right now because I'm still somewaht familiar with it and it will save you a LOT of time.

- anonymous

Thats cool.
The probability of selecting a derangement of n items is 1/e as n->oo

- anonymous

If you randomly mix an ordered list of n items, the probability of selecting the sequence such that each item is in the wrong place, as n->oo , this probability is 1/e

- wolf1728

For n=3, there are 2 possible derangements
When n=4 we use the number of derangements for n=3 which is 2
We are talking about n=4 derangements so we get
PREVIOUS successes which = 2
n=4
we multiply 4 * (PREVIOUS derangements)
=8
then we ADD 1 because we are working with an "even" n of 4
SO 4 * 2 +1 = 9
When n = 5 derangements equal
5 * (PREVIOUS derangements) MINUS 1 because n is 5 which is odd.
5 * 9 - 1
= 44 Derangements for n=5
For n = 6
6 * (PREVIOUS derangements) PLUS 1 because n=6 an EVEN number
6 * 44 + 1 which equals
265
Yes a little cumbersome but MUCH simpler than the other formula.
I discovered this formula on my own. It has NO proof - but it works!!!

- wolf1728

Oh geez now you are talking about the 1/e probability.
Anyway, you now have an EASY way for derangement calculation.
If you have any questions, just send me an E-Mail
My E-Mail address is here:
http://1728.org/

- wolf1728

Gonna get going jayzdd
Nice meeting you here.
c ya :-)

- anonymous

Thanks

- anonymous

Interesting, you found a recursive formula to produce derangements.

- anonymous

With some consideration whether n is even or odd

- anonymous

There might be a way to prove using advanced math techniques.

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