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mathmath333

  • one year ago

Find no of ways 7 boys and 7 girls can sit around a round table such that no 2 boys or 2 girls are together.

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Find no of ways 7 boys and 7 girls can sit around a round table }\hspace{.33em}\\~\\ & \normalsize \text{ such that no 2 boys or 2 girls are together.}\hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    |dw:1442401794759:dw|

  3. imqwerty
    • one year ago
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    :O

  4. mathmath333
    • one year ago
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    is it 7!*7!

  5. imqwerty
    • one year ago
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    |dw:1442401981845:dw|

  6. arindameducationusc
    • one year ago
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    I guess it is (7-1)!*7! according to the diagram you have given

  7. arindameducationusc
    • one year ago
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    but the question says 2 boys and 2 girls alternatively?

  8. mathmath333
    • one year ago
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    yes in book it is 6!*7!

  9. arindameducationusc
    • one year ago
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    my bad..... ! I didn't read the question properly, it says no two boys and girls.....

  10. mathmath333
    • one year ago
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    how does (7-1)! comes ?

  11. anonymous
    • one year ago
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    if you have n items, the number of ways to rearrange them in a circle is (n-1)! you have 14/2 pairs of boy girl , so there are 6! ways to arrange them

  12. mathmath333
    • one year ago
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    but by that logic it should be =(7-1)!*(7-1)!=6!*6!

  13. arindameducationusc
    • one year ago
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    hmmm..... no no

  14. arindameducationusc
    • one year ago
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    Its because the arrangement for the people anticlockwise and clockwise will be same hence these arrangements are considered as one because order is not changing..... hmmmm..... LET me think of a better explanation....

  15. mathmate
    • one year ago
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    You can think of it this way |dw:1442402619890:dw|

  16. mathmate
    • one year ago
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    A boy chooses any of the 14 seats, and it won't change anything because of rotation.

  17. mathmate
    • one year ago
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    Then there are 6! to seat the boys.

  18. mathmate
    • one year ago
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    The girls sit between the boys, so 7!

  19. anonymous
    • one year ago
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    first count the number of ways to rearrange pairs of bg in a circle bg , bg, bg, bg, bg , bg, bg this is 6! now for each of these circle arrangements, you can also rearrange the girls , fixing the boys. there are 7! ways to rearrange the girls

  20. anonymous
    • one year ago
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    thats a total of 6! 7!

  21. arindameducationusc
    • one year ago
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    |dw:1442402759464:dw| just assisting :D

  22. mathmath333
    • one year ago
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    ok thnks all

  23. imqwerty
    • one year ago
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    no prblm hehe :D

  24. anonymous
    • one year ago
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    |dw:1442403041001:dw|for each arrangement of boys, the girls can rearrange in 7! ways. and the boys can rearrange in 6! ways on a circle

  25. anonymous
    • one year ago
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    that is 6! * 7! total

  26. anonymous
    • one year ago
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    |dw:1442403160561:dw|

  27. mathmath333
    • one year ago
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    suppose their were 3 persons then ?

  28. anonymous
    • one year ago
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    |dw:1442403194062:dw|

  29. imqwerty
    • one year ago
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    ok see this :) - |dw:1442403062721:dw| fix the boys nd then u can arrange the grls in 7! ways now u move the boys nd then u will have 6! ways to rearrange we don't count 7! cause we already counted 1 in when we fixed the boys :) so total ways = 6!*7!

  30. anonymous
    • one year ago
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    with 3 people, you will have 2 boys or 2 girls sitting next to each other. that is pigeonhole principle

  31. mathmath333
    • one year ago
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    |dw:1442403195959:dw|

  32. imqwerty
    • one year ago
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    and what are the conditions ? like a cant sit with b or c or something like that?

  33. anonymous
    • one year ago
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    oh 3 boys and 3 girls

  34. mathmath333
    • one year ago
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    no boy, girl ,alien

  35. Jhannybean
    • one year ago
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    alien...lmao

  36. anonymous
    • one year ago
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    3 boys 3 girls 3 aliens no 2 boys, no 2 girls, no 2 aliens sit next each other?

  37. mathmath333
    • one year ago
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    ya this one ^

  38. mathmath333
    • one year ago
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    will it be 3!*3!*2!

  39. anonymous
    • one year ago
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    lets label them b1, g1, a1 b2, g2, a2 b3, g3, a3 place them on a circle fix the aliens with a thumbtack, and start counting the b's and g's

  40. mathmath333
    • one year ago
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    3!*3!*2!

  41. anonymous
    • one year ago
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    |dw:1442403829346:dw|

  42. anonymous
    • one year ago
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    |dw:1442403873463:dw|

  43. anonymous
    • one year ago
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    when you start placing girls, you have more choices

  44. mathmath333
    • one year ago
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    3!3!6!

  45. anonymous
    • one year ago
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    |dw:1442403928373:dw|

  46. anonymous
    • one year ago
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    remember we started by fixing a1,a2,a3 the aliens with a thumbtack there are 2! ways to rearrange those. for each alien there are 3! ways to place the boys. for each boy there are 6*5*4 ways to place the girls 2! * 3! * 6*5*4

  47. anonymous
    • one year ago
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    or the number of ways to rearrange xxx g1 g2 g3 is 6! / (3! * 1!*1! *1!)

  48. anonymous
    • one year ago
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    I get 1440

  49. anonymous
    • one year ago
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    'x' stands for a space a girl will not land in , its blank

  50. mathmath333
    • one year ago
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    how u get this "6*5*4".

  51. anonymous
    • one year ago
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    you saw how i placed the aliens, the boys, then we have left the girls

  52. mathmath333
    • one year ago
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    |dw:1442404331340:dw|

  53. mathmath333
    • one year ago
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    |dw:1442404352769:dw|

  54. mathmath333
    • one year ago
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    |dw:1442404370104:dw|

  55. anonymous
    • one year ago
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    |dw:1442404420852:dw|

  56. mathmath333
    • one year ago
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    what is "x" in ur diagram

  57. anonymous
    • one year ago
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    those are blank spaces, the girls do not land in them . but they are important in the configuration

  58. anonymous
    • one year ago
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    the girls can switch places with the x's

  59. mathmath333
    • one year ago
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    ok i get it now

  60. anonymous
    • one year ago
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    then the question is how many ways to rearrange g1 g2 g3 x x x

  61. anonymous
    • one year ago
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    what is interesting about theses problems, it is the first group that we 'fix' , (imagine placing thumbtacks on them), that have the circular permutations. the rest are regular or non circular permutations

  62. anonymous
    • one year ago
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    these*

  63. mathmath333
    • one year ago
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    yep i agree

  64. mathmath333
    • one year ago
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    thats why they are Counting questions

  65. mathmath333
    • one year ago
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    |dw:1442404776805:dw|this is imquerty

  66. imqwerty
    • one year ago
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    lol

  67. imqwerty
    • one year ago
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    how do u make such drawings :D

  68. mathmath333
    • one year ago
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    with tool extension

  69. mathmath333
    • one year ago
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    works in chrome

  70. imqwerty
    • one year ago
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    awww :( i experimented with chrome nd now its not wrkin i need to completely uninstall it but it doesnt wrk ;-;

  71. mathmath333
    • one year ago
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  72. mathmath333
    • one year ago
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    this is that tool ^

  73. imqwerty
    • one year ago
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    ;-; i will fix it after 2 weeks :( till then i hav to use the old tool ;-;

  74. mathmath333
    • one year ago
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    lol it works instantly

  75. anonymous
    • one year ago
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    this is how math gets deeper and deeper

  76. imqwerty
    • one year ago
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    :)

  77. anonymous
    • one year ago
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    how to solve the general problem. suppose we have x11 , x12, x13, ... x1k cannot sit next to each other x21, x22, x23 ... , x2k cannot sit next to each other ... ... xn1, xn2, xn3, ... xnk cannot sit next to each other on a circle, how many ways can they be seated.

  78. mathmath333
    • one year ago
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    this is hard wont come in ma exam

  79. anonymous
    • one year ago
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    this can be a research problem :)

  80. anonymous
    • one year ago
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    we have done some base cases

  81. mathmath333
    • one year ago
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    yea not for inocents like me

  82. anonymous
    • one year ago
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    :D

  83. mathmath333
    • one year ago
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    |dw:1442405691978:dw|

  84. anonymous
    • one year ago
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    Thank you for using the OpenStudy Notebook extension.

  85. mathmath333
    • one year ago
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    welcome

  86. anonymous
    • one year ago
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    |dw:1442406626294:dw|

  87. arindameducationusc
    • one year ago
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    can you guys give me the extension link?

  88. mathmath333
    • one year ago
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    The one i attached before will work

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