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mathmath333

  • one year ago

A teacher has to distribute 15 pens among 5 of his students such that student A gets at least 3 and at most 6 pens.In how many ways can this be done ?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{A teacher has to distribute 15 pens among 5 of his students such}\hspace{.33em}\\~\\ & \normalsize \text{that student A gets at least 3 and at most 6 pens.In how many }\hspace{.33em}\\~\\ & \normalsize \text{ ways can this be done ?}\hspace{.33em}\\~\\ & a.)\ 495 \hspace{.33em}\\~\\ & b.)\ 77 \hspace{.33em}\\~\\ & c.)\ 417 \hspace{.33em}\\~\\ & d.)\ 435 \hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    only student A gets atleast 3 to 6 pens necessarily

  3. dan815
    • one year ago
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    lets do 4 cases then Student A 3 4 5 6

  4. dan815
    • one year ago
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    when student A gets 3 pens we have 12 pens to distribute among 4 students

  5. mathmath333
    • one year ago
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    12C4=495

  6. dan815
    • one year ago
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    u cant just do that

  7. dan815
    • one year ago
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    why are you doing 12 choose 4

  8. mathmath333
    • one year ago
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    cuz i left with 4 students and 12 pens

  9. dan815
    • one year ago
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    wait umm from your previous questions, when they ask questions like this does that mean the pens are distinguishable or not?

  10. mathmath333
    • one year ago
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    i think pens are similar/identical

  11. dan815
    • one year ago
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    okay phew that makes it a lot simpler

  12. dan815
    • one year ago
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    so we have 12 pens and 4 students think about it like stars and bars question

  13. mathmath333
    • one year ago
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    15C3

  14. dan815
    • one year ago
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    |dw:1441579954826:dw|

  15. dan815
    • one year ago
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    yes 15 choose 3

  16. dan815
    • one year ago
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    now can u do the other cases?

  17. mathmath333
    • one year ago
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    14C3 ?

  18. dan815
    • one year ago
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    go on

  19. mathmath333
    • one year ago
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    15C3+14C3+12C3+11C3 ?

  20. dan815
    • one year ago
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    yeah

  21. mathmath333
    • one year ago
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    i m dumb

  22. mathmath333
    • one year ago
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    u mean it is correct

  23. mathmath333
    • one year ago
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    it is 1325 not in options

  24. dan815
    • one year ago
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    hm thats weird, i dont see anything wrong with what we did, unless the question should be interpretated differently

  25. dan815
    • one year ago
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    @Empty

  26. mathmath333
    • one year ago
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    what if the pens are distinct .

  27. beginnersmind
    • one year ago
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    You had 15 pens originally but you gave at least 3 to A.

  28. dan815
    • one year ago
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    then it iwll be even more

  29. beginnersmind
    • one year ago
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    So the highest case is 12C3

  30. beginnersmind
    • one year ago
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    and the lowest 9C3

  31. anonymous
    • one year ago
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    12C4+11C4+10C4+9C4 is wrong ?

  32. dan815
    • one year ago
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    why 12C3?

  33. mathmath333
    • one year ago
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    answer given is d.) 435

  34. mathmath333
    • one year ago
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    yea why 12C3

  35. beginnersmind
    • one year ago
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    @Halmos Yes, because you're not choosing people. You are choosing where to put the dividing line between people's stack of pens.

  36. beginnersmind
    • one year ago
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    |dw:1441580705091:dw| orginal credit to dan. You started out with 15 pens. But you gave 3 to A. So you have 12 left.

  37. dan815
    • one year ago
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    right but after u get 12 pens then u got a total of 15 spots from the 12 pens and 3 divider lines

  38. beginnersmind
    • one year ago
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    BTW 12C3 is incorrect too. It should be 11C3. Because there's 11 positions to put your dividing line.

  39. beginnersmind
    • one year ago
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    |dw:1441580893991:dw| Or 13. And you can put 2 dividing lines in the same position as well :(

  40. beginnersmind
    • one year ago
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    So it's really 3^13 +3^12 +3^11 +3^10

  41. dan815
    • one year ago
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    i reallly think its 15 choose 3 for dividing 12 pens between 4ppl if we allow someone to get 0 pens

  42. dan815
    • one year ago
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    for example 5 apples between 2 ppl so places = 5 apples + 1 division = 6 places you have 6 places to decide where your division is 6 choose 1 = 6 ways

  43. dan815
    • one year ago
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    which is true 0 ,5 1, 4 2, 3 3,2 4,1 5, 0

  44. triciaal
    • one year ago
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    student A at least 3 pens 3 or 4 or 5 or 6 next student 12 for 4 students |dw:1441581202316:dw|

  45. beginnersmind
    • one year ago
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    So how do you get 15 places from 12 objects?

  46. dan815
    • one year ago
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    3 for divisions

  47. dan815
    • one year ago
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    what we are about is the order of these 15 things, that determines how its been split up

  48. dan815
    • one year ago
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    like if u have 1,1,1,|,| ----- 3 of 1s and 2 bars, the number of distinguishable ways to arrange this will tell you how many ways to distribute 3 identical objects for 3 people

  49. dan815
    • one year ago
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    so in this case u have 5 spots for your 2 bars, 5C2

  50. beginnersmind
    • one year ago
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    You are confounding choosing positions for your dividers and arranging objects in a certain order.

  51. triciaal
    • one year ago
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    |dw:1441581658688:dw|

  52. beginnersmind
    • one year ago
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    The way you should think about it is that you have 3 dividers and 13 positions to put the in. More than 1 divider can go in the same position but all the dividers are indistinguishable.

  53. dan815
    • one year ago
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    yes that is right

  54. dan815
    • one year ago
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    well instead of positions and stuff

  55. dan815
    • one year ago
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    you can think about it like 2 distinguishable objets 12 of 1, 3 of the other, and the order they are placed in will determine the unique solutions

  56. beginnersmind
    • one year ago
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    That's another way to do it, yes. They should give the same answer. Mine is 3^13, yours is 15!/(12!*3!)

  57. beginnersmind
    • one year ago
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    Correction Mine is (3^13)/3!, yours is 15!/(12!*3!)

  58. beginnersmind
    • one year ago
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    15!/(12!*3!) = 15C3 Mind Blown

  59. triciaal
    • one year ago
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    |dw:1441581910685:dw|

  60. beginnersmind
    • one year ago
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    Ok, @dan815, finally figured it out. Your method works, mine doesn't. Sorry, I'm slow tonight.

  61. dan815
    • one year ago
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    ah its all good, so why do u think its not in the answers though

  62. beginnersmind
    • one year ago
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    Not sure yet.

  63. beginnersmind
    • one year ago
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    I'm trying to figure it if there's any double counting going on but I can't see any.

  64. beginnersmind
    • one year ago
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    Alright, I give up. Good luck everyone. @mathmath33 you might want to ask your teacher.

  65. triciaal
    • one year ago
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    isn't Student A one particular student of the 5? if a student then you have multiples

  66. triciaal
    • one year ago
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    or like @beginnersmind calls it "doublecounting"

  67. beginnersmind
    • one year ago
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    Well, here's two different related problems. You can try to solve them hoping that one of them gives an answer that's one of the options. Version1: A teacher has to distribute 15 pens among 5 of his students such that there is at least one student who get at least 3 and at most 6 pens.In how many ways can this be done ? Version2: A teacher has to distribute 15 pens among 5 of his students such that there is exactly one student who gets at least 3 and at most 6 pens.In how many ways can this be done ? Version3: A teacher has to distribute 15 pens among 5 of his students such that every student gets at least 3 and at most 6 pens.In how many ways can this be done ? Hint: Version 3 is the easiest :)

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