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Ways to put letters in postboxes

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  1. Jul 31, 2015 #1
    1. The problem statement, all variables and given/known data

    There are 7 different postbox, and 10 identical letters. How many ways can the letters put into the boxes so that there is at least one letter in a postbox?

    2. Relevant equations
    nCr=n!/(n-r)!r!

    If M,N,O..... things can be done in m,n,o..... ways then ways of doing them together is m*n*o..


    3. The attempt at a solution

    First I tried to put 7 letters in 7 boxes. It can be done in 1 way only as all the letters are identical. Then I tried to put the rest three letters in the box. Let us mark the letters as A, B and C (just to describe easily). While putting A in a box, for every box there are two solutions, either yes you put or no you don't put. So ways are 2^7. But it also includes that you dont put it any boxes. So, actual ways are 2^7-1. Same for letter B and C. So total ways can be (2^7-1)^3 = 2048383 ways.

    The problem solution was given a number something less than 100. I think the first step maybe a problem where I said ways to put 7 letters is 1. But then the answer is going to be bigger, not less than 100 anyways. I asked the one who gave this problem about my solution, he couldn't answer. It was on facebook and not on my id. And I don't remember how his solution was. Can anybody tell me where the problem is?
     
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  3. Jul 31, 2015 #2

    Borg

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    Since you have to have one letter in each box, the number of combinations is the same as the number of ways to arrange 3 idential letters in 10 different boxes. Does that make it easier for you?
     
  4. Jul 31, 2015 #3
    Yes, but how to put these three in 7 boxes (Maybe you mistook 10)? Is my second step to solve this a good approach?
     
  5. Jul 31, 2015 #4
    Maybe the solution given there was

    1) All three letters in same box in 7C1

    2) 2 letters in a box and rest in another box in 7C1*6C1

    3) 3 letters in 3 different boxes in 7C3

    Then sum of case 1, 2 and 3.

    This makes sense to me. But what's wrong with my concept then?
     
  6. Jul 31, 2015 #5

    RUber

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    Fireflies, in your original method, you not only are allowing for no box to get a letter from the leftover 3, but also for every box to get a letter (7>3). That is why your solution is so large.
    --As you just posted--
    You have the options to put:
    1 additional letter into 3 boxes
    2 additional letters into 1 box and 1 letter into another
    3 additional letters into 1 box
    This is a good way to look at the problem.
     
  7. Jul 31, 2015 #6

    Borg

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    Yes, I misread. With 3 letters spread across 7 boxes.

    The case that you have here is if all of the boxes are identical.
    Yech, another misread.
     
    Last edited: Jul 31, 2015
  8. Jul 31, 2015 #7
    Yes, all the boxes are not identical. That's my point.


    Well, I subtracted that 1 to confirm no box is selected. But that includes all yes options too. Now I get what's the actual mistake hete
     
  9. Jul 31, 2015 #8
    Then if I say in this way, can this be right-
    For letter A to select a box is 7C1. Same for letter B and letter C? So, finally (7C1)^3

    Or it is wrong as the letters are identical?
     
  10. Jul 31, 2015 #9

    Borg

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    Since the letters are identical, there are only 7 ways to fill a box with 3 letters. Does that make sense?
     
  11. Jul 31, 2015 #10

    RUber

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    That would be wrong, since with identical letters AC = AB, so 7^3 includes combinations with AC, AB, BC in the same box counted as distinct outcomes.
     
  12. Jul 31, 2015 #11
    So, you are saying that the way with three cases (in #4) is correct? What would be the case then if the post-boxes were identical?
     
  13. Jul 31, 2015 #12
    No, how? The post boxes are not identical. 3 letters in each box itself is going to be 7 ways. 2 in one box rest in other can add more choices like box no 1,2; 2,1; 1,3 etc. 3rd case will give extra choices.
     
  14. Jul 31, 2015 #13

    Borg

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    Post #5 gives you the 3 variations that you have to consider.
    Post #9 gives you the answer to one of those variations.
     
  15. Jul 31, 2015 #14
    Sorry, maybe I got it wrong. All three letters in a box and this can be done in 7 ways, that does make sense
     
  16. Jul 31, 2015 #15
    Now I can actually get what you're trying to say :)

    But I already did that in post#4 right?
     
  17. Jul 31, 2015 #16

    RUber

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    If the boxes were identical and the letters were identical you would only have:
    6 boxes with 1, 1 box with 4
    5 boxes with 1, 1 box with 2, 1 box with 3
    4 boxes with 1, 3 boxes with 2.
    That is: 3 ways to split 10 letters into 7 boxes with at least one in each box.

    This is almost certainly not what the original question was looking for.
     
  18. Jul 31, 2015 #17
    Since now it is more clear to me, let me try to answer post#11 by myself. Here post-boxes are identical. So, we can only do is
    3 in a box in 3C3
    2 in a box and rest in other in 3C2
    1 in each box in 3C1,

    Then add all. Am I correct?
     
  19. Jul 31, 2015 #18
    I read the post later than I posted my answer. It makes me confused again. Why not we put the 7 in their case and try to combine the rest three here?

    Yes, it is not an answer to the original post, but since I am having problems to work out the conditions so I think these answers are more likely to help me how to solve these actually
     
  20. Jul 31, 2015 #19

    Borg

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    You have the answer to this one. So let's look at the next one up:
    Just answer this question first: How many ways can you put 2 additional letters into 1 box?
     
  21. Jul 31, 2015 #20

    RUber

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    Are you still assuming letters are identical too?

    3 in a box means 3 in one box 0 in others. If boxes are identical, there is only one way to do it.
    2 in one box, 1 in another, and 0 in others. This is still one case, all similar layouts are considered indistinct.
    1 in each of 3 boxes and 4 empty boxes. Also, just one case.

    If the letters are not identical, then you need to start from 10 letters again. I reiterate, this is not what the question was asking and would likely become complicated quickly.
    ( I could be wrong on these since I don't often deal with these sorts of questions)
    Case 1: 4 in one box, one in each of the other 6-- 10C4 ways to choose the 4 letters that go in.
    Case 2: 3 in one box, two in another, 1 in rest--10C5 ways to choose the ones going into two boxes times 5C3 ways to choose which ones go into the first.
    Case 3: 2 in three boxes, 1 in the rest -- 10C6 * 6C4 * 4C2 ways

    Then sum the 3 cases.
     
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