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Ehsan18
 one year ago
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
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0suppose 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
 one year ago
Best ResponseYou've already chosen the best response.2In 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! * (11/1! +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
 one year ago
Best ResponseYou've already chosen the best response.2LOL jayzdd gets the medal for an INCORRECT answer ??? !!! The answer is NINE!! LOL

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Heck, I'LL send a medal to jayzdd too!!!

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Well 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
 one year ago
Best ResponseYou've already chosen the best response.0Using 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
 one year ago
Best ResponseYou've already chosen the best response.0Nice derangement formula. I did this by making a tree which is tedious

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Hello to ganeshie8 :)

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Or you could have just copied my numbers :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434277905759:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0some of these branches won't work, because we are left with a 4 in the last branch

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2and anyway, to continue with your ABCD method the ones remaining are: BDAC CADB CDBA DCAB

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434278137565:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I like the tree way because you can use your fingers to cover the possibilities already covered , systematically.

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Okay Jayzdd you stuck it out with the "tree" method. I admire your persistence. That is a good trait.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thankyou . we came out with the same exact solution. Maybe your trial and error was similar?

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Yes. 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
 one year ago
Best ResponseYou've already chosen the best response.2And scrolling down, the scroll bar stops about half way down.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sometimes it gets laggy because of cache or scripts running in background . You might have to restart your browser, thats what I usually.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you know how they derived that derangements formula?

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Hmm 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
 one year ago
Best ResponseYou've already chosen the best response.0Thats your website . Nice

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Shucks 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.

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Heck I bet you never knew about "derangements" before tonight right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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!  (

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Heck, 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
 one year ago
Best ResponseYou've already chosen the best response.0its nine I get it.....Thanks for the answer

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Hey 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
 one year ago
Best ResponseYou've already chosen the best response.2And yes ehsan18 it is nine derangements and you'll probably see about a dozen explanations why it is nine LOL

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If i follow that , that will explain the process?

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0Thats cool. The probability of selecting a derangement of n items is 1/e as n>oo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.2For 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
 one year ago
Best ResponseYou've already chosen the best response.2Oh 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 EMail My EMail address is here: http://1728.org/

wolf1728
 one year ago
Best ResponseYou've already chosen the best response.2Gonna get going jayzdd Nice meeting you here. c ya :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Interesting, you found a recursive formula to produce derangements.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0With some consideration whether n is even or odd

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There might be a way to prove using advanced math techniques.
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