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mathmath333

  • one year ago

Counting Problem

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{4 men and 3 women are to be seated in row so that } \hspace{.33em}\\~\\ & \normalsize \text{no two women sit together. } \hspace{.33em}\\~\\ & \normalsize \text{Find the number of ways in which they can be seated. } \hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    ok so we lsit the ways to do this can you think of any

  3. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & a.)\ 4!\times ^{5}P_{3} \hspace{.33em}\\~\\ & b.)\ 4!\times ^{5}C_{3} \hspace{.33em}\\~\\ & c.)\ 4!\times ^{4}P_{3} \hspace{.33em}\\~\\ & d.)\ 4!\times ^{4}C_{3} \hspace{.33em}\\~\\ \end{align}}\)

  4. mathmath333
    • one year ago
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    @mathmate

  5. mathmath333
    • one year ago
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    i have found a link can any one explan me http://www.meritnation.com/ask-answer/question/m-men-and-n-women-are-to-be-seated-in-a-row-so-that-no-two-w/permutations-and-combinations/5901983

  6. mathmate
    • one year ago
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    Yes, it does help!

  7. mathmate
    • one year ago
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    So we first seat the 4 men, so there are 4! ways of seating the gents. |dw:1440446507784:dw|

  8. dan815
    • one year ago
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    yep that pic explains it perfectly

  9. mathmate
    • one year ago
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    Now we leave 5= (4+1) chairs between and beside the gents for the three women to choose. So there are 5P3 ways to sit them.

  10. dan815
    • one year ago
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    ill give u another question similiar to this , try to figure it out :)

  11. mathmate
    • one year ago
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    I'll try! lol

  12. dan815
    • one year ago
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    what if the 4 men and 3 women are seated around a round table

  13. mathmate
    • one year ago
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    with the same constraint?

  14. dan815
    • one year ago
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    yes

  15. mathmate
    • one year ago
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    lemme see.

  16. mathmath333
    • one year ago
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    0 ways

  17. dan815
    • one year ago
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    haha

  18. dan815
    • one year ago
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    ya

  19. mathmate
    • one year ago
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    It's cheaper!

  20. mathmath333
    • one year ago
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    is 0 correct

  21. dan815
    • one year ago
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    ya its impossible ofcourse lol

  22. mathmath333
    • one year ago
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    yay!

  23. dan815
    • one year ago
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    |dw:1440447006885:dw|

  24. mathmate
    • one year ago
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    |dw:1440446969033:dw| So for the four men, there are 3! ways (rotational symmetry)

  25. mathmate
    • one year ago
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    For the women, there are only 4 seats, so 4P3. Now we multiply them together!

  26. mathmate
    • one year ago
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    For some reason, men have to fight for women! lol

  27. dan815
    • one year ago
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    remember its not 4 men and 4 women mathmate,

  28. mathmath333
    • one year ago
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    yes^ i was wondering that pic

  29. mathmate
    • one year ago
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    Four seat, but not all filled!

  30. mathmate
    • one year ago
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    First women gets to choose from 4, then 3, then 2.

  31. mathmate
    • one year ago
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    *woman

  32. mathmate
    • one year ago
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    One seat will be empty, so remove the chair and two men will sit next to each other.

  33. dan815
    • one year ago
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    yep

  34. dan815
    • one year ago
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    here is a more involved question

  35. mathmate
    • one year ago
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    Woah, not too difficult, I hope!

  36. dan815
    • one year ago
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    4 women and 4 men how many ways to ARRANGE them in a row, where 2 women cannot sit beside each other. How many ways to Arrange them around a round table,(note*the seat number is not what the counts but the pattern of the arrangement*)

  37. mathmate
    • one year ago
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    Is this a trick question? They seem like the ones we did.

  38. dan815
    • one year ago
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    no this one is more straightforward but theres some nice ways to do it efficiently

  39. dan815
    • one year ago
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    ill leave u the answers see if u get them both

  40. mathmate
    • one year ago
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    4!5P4 for the first one, and 3!4P4 for the second.

  41. dan815
    • one year ago
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    part a =1152 part b= 72

  42. mathmate
    • one year ago
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    So I got them both wrong! lol

  43. mathmath333
    • one year ago
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    for first question of round table is this possible|dw:1440447723500:dw|

  44. dan815
    • one year ago
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    oh oops ya, i thought the question said no 2 women and no 2 wmen can sit together

  45. mathmath333
    • one year ago
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    i refuted my answer lol

  46. dan815
    • one year ago
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    no i wanted it to be like that, no gender together xD

  47. mathmath333
    • one year ago
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    oh it is different

  48. mathmate
    • one year ago
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    Yea, for the first one (4 gents+4 ladies), we put the women on the men's right, so there are 4!4!=576 ways. Now put the ladies on the left of the gents, we get another 576, that makes 1152.

  49. dan815
    • one year ago
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    yep thats right

  50. dan815
    • one year ago
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    ill check your first question answer in the mean time

  51. mathmate
    • one year ago
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    For the second, I still get 144=3!4!. 3! for the men with rotational symmetry. Once the men are seated, there are 4 distinct seats for the 4 ladies, so 4!

  52. dan815
    • one year ago
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    u r close

  53. dan815
    • one year ago
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    remember that now since its m and w on a round table mwmwmw = wmwmwm

  54. dan815
    • one year ago
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    so its not 1152/8 but 1152/2/8

  55. mathmate
    • one year ago
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    Oh, for the straight row, I have 1152=2*4!4! But for the round table, I still have 144=3!4!

  56. dan815
    • one year ago
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    well the way i connected the first answer to the 2nd is

  57. dan815
    • one year ago
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    u see how its 2*4!*4! theres no need for the 2 * anymore in the 2nd case since

  58. dan815
    • one year ago
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    w1m1w2m2 = m1w1m2w2 if u connect the edges its the same arrangement

  59. dan815
    • one year ago
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    so u have 4!*4! now also since its a round table u can keep shifting every arrangement by 8 times to get the same one around the round table

  60. dan815
    • one year ago
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    w1m1w2m2 for example this is equal to m2w1m1w2

  61. dan815
    • one year ago
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    |dw:1440448675569:dw|

  62. dan815
    • one year ago
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    so for 8 seats there are 8 repeated arranements that are coming out of the row case when u move to the round table case

  63. mathmate
    • one year ago
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    That's true, I forgot the men can sit in the women's seats at the very start. I place the men, then throw in the chairs! lol I get it now, 72 is the number!

  64. dan815
    • one year ago
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    yep :)

  65. dan815
    • one year ago
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    do u get it too mathmath333

  66. mathmate
    • one year ago
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    Sorry, @mathmath333, we hijacked your post! :(

  67. dan815
    • one year ago
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    hehe

  68. mathmath333
    • one year ago
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    was offline ,lol

  69. mathmate
    • one year ago
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    Sorry mm333, we were discussing extensions to your problem! lol

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