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?

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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?

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

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Heck, I'LL send a medal to jayzdd too!!!
I did not finish
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?)
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
Nice derangement formula. I did this by making a tree which is tedious
Hello to ganeshie8 :-)
Or you could have just copied my numbers :-)
|dw:1434277905759:dw|
some of these branches won't work, because we are left with a 4 in the last branch
and anyway, to continue with your ABCD method the ones remaining are: BDAC CADB CDBA DCAB
|dw:1434278137565:dw|
I like the tree way because you can use your fingers to cover the possibilities already covered , systematically.
Okay Jayzdd you stuck it out with the "tree" method. I admire your persistence. That is a good trait.
Thankyou . we came out with the same exact solution. Maybe your trial and error was similar?
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.
And scrolling down, the scroll bar stops about half way down.
Sometimes it gets laggy because of cache or scripts running in background . You might have to restart your browser, thats what I usually.
do*
Do you know how they derived that derangements formula?
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.
Thats your website . Nice
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.
oh :)
Heck I bet you never knew about "derangements" before tonight right?
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! - (
I'll research this.
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.
its nine I get it.....Thanks for the answer
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.
And yes ehsan18 it is nine derangements and you'll probably see about a dozen explanations why it is nine LOL
If i follow that , that will explain the process?
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.
Thats cool. The probability of selecting a derangement of n items is 1/e as n->oo
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
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!!!
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/
Gonna get going jayzdd Nice meeting you here. c ya :-)
Thanks
Interesting, you found a recursive formula to produce derangements.
With some consideration whether n is even or odd
There might be a way to prove using advanced math techniques.

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