## 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.

1. mathmath333

\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

|dw:1442401794759:dw|

3. imqwerty

:O

4. mathmath333

is it 7!*7!

5. imqwerty

|dw:1442401981845:dw|

6. arindameducationusc

I guess it is (7-1)!*7! according to the diagram you have given

7. arindameducationusc

but the question says 2 boys and 2 girls alternatively?

8. mathmath333

yes in book it is 6!*7!

9. arindameducationusc

my bad..... ! I didn't read the question properly, it says no two boys and girls.....

10. mathmath333

how does (7-1)! comes ?

11. anonymous

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

but by that logic it should be =(7-1)!*(7-1)!=6!*6!

13. arindameducationusc

hmmm..... no no

14. arindameducationusc

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

You can think of it this way |dw:1442402619890:dw|

16. mathmate

A boy chooses any of the 14 seats, and it won't change anything because of rotation.

17. mathmate

Then there are 6! to seat the boys.

18. mathmate

The girls sit between the boys, so 7!

19. anonymous

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

thats a total of 6! 7!

21. arindameducationusc

|dw:1442402759464:dw| just assisting :D

22. mathmath333

ok thnks all

23. imqwerty

no prblm hehe :D

24. anonymous

|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

that is 6! * 7! total

26. anonymous

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27. mathmath333

suppose their were 3 persons then ?

28. anonymous

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29. imqwerty

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

with 3 people, you will have 2 boys or 2 girls sitting next to each other. that is pigeonhole principle

31. mathmath333

|dw:1442403195959:dw|

32. imqwerty

and what are the conditions ? like a cant sit with b or c or something like that?

33. anonymous

oh 3 boys and 3 girls

34. mathmath333

no boy, girl ,alien

35. Jhannybean

alien...lmao

36. anonymous

3 boys 3 girls 3 aliens no 2 boys, no 2 girls, no 2 aliens sit next each other?

37. mathmath333

ya this one ^

38. mathmath333

will it be 3!*3!*2!

39. anonymous

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

3!*3!*2!

41. anonymous

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42. anonymous

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43. anonymous

when you start placing girls, you have more choices

44. mathmath333

3!3!6!

45. anonymous

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46. anonymous

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

or the number of ways to rearrange xxx g1 g2 g3 is 6! / (3! * 1!*1! *1!)

48. anonymous

I get 1440

49. anonymous

'x' stands for a space a girl will not land in , its blank

50. mathmath333

how u get this "6*5*4".

51. anonymous

you saw how i placed the aliens, the boys, then we have left the girls

52. mathmath333

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53. mathmath333

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54. mathmath333

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55. anonymous

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56. mathmath333

what is "x" in ur diagram

57. anonymous

those are blank spaces, the girls do not land in them . but they are important in the configuration

58. anonymous

the girls can switch places with the x's

59. mathmath333

ok i get it now

60. anonymous

then the question is how many ways to rearrange g1 g2 g3 x x x

61. anonymous

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

these*

63. mathmath333

yep i agree

64. mathmath333

thats why they are Counting questions

65. mathmath333

|dw:1442404776805:dw|this is imquerty

66. imqwerty

lol

67. imqwerty

how do u make such drawings :D

68. mathmath333

with tool extension

69. mathmath333

works in chrome

70. imqwerty

awww :( i experimented with chrome nd now its not wrkin i need to completely uninstall it but it doesnt wrk ;-;

71. mathmath333

72. mathmath333

this is that tool ^

73. imqwerty

;-; i will fix it after 2 weeks :( till then i hav to use the old tool ;-;

74. mathmath333

lol it works instantly

75. anonymous

this is how math gets deeper and deeper

76. imqwerty

:)

77. anonymous

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

this is hard wont come in ma exam

79. anonymous

this can be a research problem :)

80. anonymous

we have done some base cases

81. mathmath333

yea not for inocents like me

82. anonymous

:D

83. mathmath333

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84. anonymous

Thank you for using the OpenStudy Notebook extension.

85. mathmath333

welcome

86. anonymous

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87. arindameducationusc

can you guys give me the extension link?

88. mathmath333

The one i attached before will work