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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?

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  1. anonymous
    • one year ago
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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...

  2. wolf1728
    • one year ago
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    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

  3. wolf1728
    • one year ago
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    LOL jayzdd gets the medal for an INCORRECT answer ??? !!! The answer is NINE!! LOL

  4. wolf1728
    • one year ago
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    Heck, I'LL send a medal to jayzdd too!!!

  5. anonymous
    • one year ago
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    I did not finish

  6. wolf1728
    • one year ago
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    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?)

  7. anonymous
    • one year ago
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    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

  8. anonymous
    • one year ago
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    Nice derangement formula. I did this by making a tree which is tedious

  9. wolf1728
    • one year ago
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    Hello to ganeshie8 :-)

  10. wolf1728
    • one year ago
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    Or you could have just copied my numbers :-)

  11. anonymous
    • one year ago
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    |dw:1434277905759:dw|

  12. anonymous
    • one year ago
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    some of these branches won't work, because we are left with a 4 in the last branch

  13. wolf1728
    • one year ago
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    and anyway, to continue with your ABCD method the ones remaining are: BDAC CADB CDBA DCAB

  14. anonymous
    • one year ago
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    |dw:1434278137565:dw|

  15. anonymous
    • one year ago
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    I like the tree way because you can use your fingers to cover the possibilities already covered , systematically.

  16. wolf1728
    • one year ago
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    Okay Jayzdd you stuck it out with the "tree" method. I admire your persistence. That is a good trait.

  17. anonymous
    • one year ago
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    Thankyou . we came out with the same exact solution. Maybe your trial and error was similar?

  18. wolf1728
    • one year ago
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    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.

  19. wolf1728
    • one year ago
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    And scrolling down, the scroll bar stops about half way down.

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

  21. anonymous
    • one year ago
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    do*

  22. anonymous
    • one year ago
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    Do you know how they derived that derangements formula?

  23. wolf1728
    • one year ago
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    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.

  24. anonymous
    • one year ago
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    Thats your website . Nice

  25. wolf1728
    • one year ago
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    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.

  26. anonymous
    • one year ago
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    oh :)

  27. wolf1728
    • one year ago
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    Heck I bet you never knew about "derangements" before tonight right?

  28. anonymous
    • one year ago
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    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! - (

  29. anonymous
    • one year ago
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    I'll research this.

  30. wolf1728
    • one year ago
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    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.

  31. Ehsan18
    • one year ago
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    its nine I get it.....Thanks for the answer

  32. wolf1728
    • one year ago
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    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.

  33. wolf1728
    • one year ago
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    And yes ehsan18 it is nine derangements and you'll probably see about a dozen explanations why it is nine LOL

  34. anonymous
    • one year ago
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    If i follow that , that will explain the process?

  35. wolf1728
    • one year ago
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    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.

  36. anonymous
    • one year ago
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    Thats cool. The probability of selecting a derangement of n items is 1/e as n->oo

  37. anonymous
    • one year ago
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    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

  38. wolf1728
    • one year ago
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    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!!!

  39. wolf1728
    • one year ago
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    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/

  40. wolf1728
    • one year ago
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    Gonna get going jayzdd Nice meeting you here. c ya :-)

  41. anonymous
    • one year ago
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    Thanks

  42. anonymous
    • one year ago
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    Interesting, you found a recursive formula to produce derangements.

  43. anonymous
    • one year ago
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    With some consideration whether n is even or odd

  44. anonymous
    • one year ago
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    There might be a way to prove using advanced math techniques.

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