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mathmath333

  • one year ago

7 different objects must be divided among 3 people . In how many ways can this be done if at least one of them gets exactly one object ?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{7 different objects must be divided among 3 people .}\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can this be done if at least one}\hspace{.33em}\\~\\ & \normalsize \text{of them gets exactly one object ?} \hspace{.33em}\\~\\ & a.)\ 2484 \hspace{.33em}\\~\\ & b.)\ 1218 \hspace{.33em}\\~\\ & c.)\ 729 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \end{align}}\)

  2. anonymous
    • one year ago
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    i know hw to do it but u do know the detail solution for this comes if u google u want me to give the link and if u have question i ll answer :)

  3. anonymous
    • one year ago
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    http://gmatclub.com/forum/1-7-different-objects-must-be-divided-among-3-people-in-how-7409.html

  4. arindameducationusc
    • one year ago
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    Combination question?

  5. mathmath333
    • one year ago
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    well the fact is that i didnt understand the quora answer

  6. imqwerty
    • one year ago
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    nd what about the gmat solution :)

  7. mathmath333
    • one year ago
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    also the gmat :)

  8. arindameducationusc
    • one year ago
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    @phi

  9. mathmath333
    • one year ago
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    isnt he offline

  10. arindameducationusc
    • one year ago
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    yes, but if he comes online, he will definitely see, Hez good!

  11. arindameducationusc
    • one year ago
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    I am trying....

  12. imqwerty
    • one year ago
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    ok 1st can u find the number of ways in which u can distribute the gifts :)consider all cases

  13. mathmath333
    • one year ago
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    7P3 ?

  14. imqwerty
    • one year ago
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    did u count the possibility in which they don't even get a single gift

  15. mathmath333
    • one year ago
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    how to do that

  16. imqwerty
    • one year ago
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    m also thinkin that

  17. arindameducationusc
    • one year ago
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    :)

  18. imqwerty
    • one year ago
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    (:

  19. mathmate
    • one year ago
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    I have the following calculation, but have succeeded in making a combinatorial verification yet. Ways to distribute 7 different objects to 3 different bins = 3^7 (order is not important in each bin) ways to choose 2 bins out of 3 = 3 Ways to distribute 7 different objects to 2 different bins = 2^7 ways to choose 1 bin out of 3 = 3 ways to distribute 7 different objects to 1 bin = 1^7 So total number of ways to distribute 7 different objects to 3 different bins, with at least one object in each bin = 3^7-3*2^7-3*1^7 = 2187-3(128)-3*(1) =1800

  20. mathmate
    • one year ago
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    *have not succeeded

  21. mathmate
    • one year ago
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    Oh, I did not read correct, it says at least one of them gets \(exactly\) one object. I will do a variation of the above and come back.

  22. ParthKohli
    • one year ago
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    Hey, I guess I know this. But I'm not sure if we're allowed to give zero objects to any of them. It depends on the case.

  23. mathmath333
    • one year ago
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    yes u r allowed to give 0 objects i think.

  24. ParthKohli
    • one year ago
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    First, we fix the person who gets one gift. That makes us fulfil the initial condition. There are \(3\) ways to choose the person who gets one gift. Now there are six gifts left as we've given one gift to one person.\[x+y = 6\]has solutions \((0,6), (1,5),(2,4),(3,3), (4,2),(5,1),(6,0) \).

  25. mathmath333
    • one year ago
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    ok

  26. ParthKohli
    • one year ago
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    You know what, there's another way to do this.

  27. mathmath333
    • one year ago
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    well i just need an understandable way lol

  28. ParthKohli
    • one year ago
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    OK, I'd done it but the net got disconnected.

  29. ParthKohli
    • one year ago
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    First consider the case where two people get one object each. First we have to choose two people, in \(\binom{3}{2}\) ways. Then we have to choose 2 objects out of 7 and permute them among these two guys. That can be done in 7P2 ways.

  30. mathmath333
    • one year ago
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    3*(7*(2^6-6))

  31. ParthKohli
    • one year ago
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    Then consider the case where only one guy gets one object. First choose the guy who gets that one object. \(3\) ways. Then the other two guys have (0,6) (2,4) (3,3) (4,2) (6,0)

  32. ParthKohli
    • one year ago
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    For the first and the last case, we can count it in this way: \[2 \times \binom{7}{6}\]For the second and the second-to-last case, here's how:\[2 \times \binom{7}{4}\times \binom{3}{2}\]For the third case,\[2\times \binom{7}{3}\times \binom{4}{3}\]

  33. ParthKohli
    • one year ago
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    Heyyyyy never mind all that, you know

  34. ParthKohli
    • one year ago
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    OK, unfortunately that's it.

  35. ParthKohli
    • one year ago
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    Have we double-counted?

  36. ParthKohli
    • one year ago
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    @ganeshie8 our saviour

  37. ParthKohli
    • one year ago
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    Case 1: Exactly two persons get exactly one object.\[\binom{3}{2}\cdot \binom{7}{2}\cdot 2 \]Case 2: Exactly one person gets exactly one object. Number of ways to choose that person is 3. Now the other two are selected.\[0 ~\& ~6: 2! \cdot \binom{7}{6} \]\[2 ~ \& ~ 4: 2! \cdot \binom{7}{4}\cdot \binom{3}{2}\]\[3 ~\& ~3:\binom{7}{3}\cdot \binom{4}{3 } \]

  38. mathmath333
    • one year ago
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    answer given =1218

  39. ParthKohli
    • one year ago
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    Still not correct.

  40. ParthKohli
    • one year ago
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    ooo got it

  41. ganeshie8
    • one year ago
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    1) Give "one" person "one" gift and forget him. This can be done in \(3\times 7\) ways. (3 ways to choose 1 person, 7 ways to choose 1 gift) 2) Then the task is to distribute "6" gifts to "2" people such that nobody gets exactly "1" gift : \(2^6-6\) Multiply them for the final answer.

  42. ganeshie8
    • one year ago
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    At step2, above, the expression \(2^6\) comes from the fact that each gift has \(2\) possible states

  43. mathmath333
    • one year ago
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    that looks simple now

  44. ParthKohli
    • one year ago
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    Much sleeker...

  45. ParthKohli
    • one year ago
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    Haha, looks like mathmath had posted that expression earlier. :P

  46. ganeshie8
    • one year ago
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    Actually it was I who posted it from mathmath's computer..

  47. ParthKohli
    • one year ago
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    ...what

  48. mathmath333
    • one year ago
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    He is a hacker

  49. ParthKohli
    • one year ago
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    oh ew i missed the sarcasm there >_<

  50. ganeshie8
    • one year ago
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    lol we were doing teamviewer earlier but my headset won't cooperate suddenly... so..

  51. ParthKohli
    • one year ago
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    Why did you subtract six?

  52. mathmath333
    • one year ago
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    to substract the possibltties where a person gets exactly 1 gift

  53. ganeshie8
    • one year ago
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    At step2, we have 6 gifts and two people. we don't want to give exactly "1" gift to anyone. Actually, we need subtract \(2*6\), I think. Need to think more...

  54. ParthKohli
    • one year ago
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    Yeah, I mean you can give exactly two gifts to two people, as long as the last guy gets five.

  55. ParthKohli
    • one year ago
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    what I mean is you can give exactly one gift to two people

  56. ganeshie8
    • one year ago
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    Yep, so the answer is wrong. It should be : \(3*7(2^6-2*6) = 1092\)

  57. ganeshie8
    • one year ago
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    because, at step2, we have 6 gifts and 2 people each person can get "1" gift in 6 ways, so 2*6 total invalid states

  58. ParthKohli
    • one year ago
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    What am I even saying ugh

  59. ganeshie8
    • one year ago
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    No, keep going..

  60. ParthKohli
    • one year ago
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    No, there are six invalid states only!

  61. ParthKohli
    • one year ago
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    But there are two people, so you're right.

  62. ParthKohli
    • one year ago
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    But why exactly is (1, 1, 5) being counted as invalid?

  63. ganeshie8
    • one year ago
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    Because the question says so : "exactly one person gets one gift"

  64. ParthKohli
    • one year ago
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    no, it says at least one person gets exactly one gift

  65. ParthKohli
    • one year ago
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    which is why we need cases

  66. ganeshie8
    • one year ago
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    Oops! I read the problem wrong, wait

  67. ParthKohli
    • one year ago
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    Hey, I found this problem's solution. http://prntscr.com/8k0bq7

  68. ParthKohli
    • one year ago
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    The solution is pretty much how I did it. The only difference was that this solution first distributed one gift to the guy who had one gift, then distributed it to another guy. What I did was that I distributed that one gift in the end.

  69. ParthKohli
    • one year ago
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    http://prntscr.com/8k0czp

  70. ganeshie8
    • one year ago
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    that is nice

  71. ParthKohli
    • one year ago
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    \[\binom{3}{2}\cdot \binom{7}{2}\cdot 2\]\[+ ~3 \left(2!\cdot \binom{7}{6} + 2!\cdot \binom{7}{4}\cdot \binom{3}{2} + \binom{7}{3}\cdot \binom{4}{3}\right)\]

  72. ganeshie8
    • one year ago
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    Hmm, wonder how is that related to \(3*7(2^6-6)\)

  73. ParthKohli
    • one year ago
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    oh that's actually right... I evaluated it wrongly the last time

  74. ParthKohli
    • one year ago
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    thanks wolfram

  75. ParthKohli
    • one year ago
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    i was getting 1638 the last time because I forgot there was no 2! for the very last term -_-

  76. ParthKohli
    • one year ago
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    what were you guys doing on teamviewer anyway

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