## mathmath333 one year ago Counting Problem

1. mathmath333

\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

ok so we lsit the ways to do this can you think of any

3. mathmath333

\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

@mathmate

5. mathmath333

6. mathmate

Yes, it does help!

7. mathmate

So we first seat the 4 men, so there are 4! ways of seating the gents. |dw:1440446507784:dw|

8. dan815

yep that pic explains it perfectly

9. mathmate

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

ill give u another question similiar to this , try to figure it out :)

11. mathmate

I'll try! lol

12. dan815

what if the 4 men and 3 women are seated around a round table

13. mathmate

with the same constraint?

14. dan815

yes

15. mathmate

lemme see.

16. mathmath333

0 ways

17. dan815

haha

18. dan815

ya

19. mathmate

It's cheaper!

20. mathmath333

is 0 correct

21. dan815

ya its impossible ofcourse lol

22. mathmath333

yay!

23. dan815

|dw:1440447006885:dw|

24. mathmate

|dw:1440446969033:dw| So for the four men, there are 3! ways (rotational symmetry)

25. mathmate

For the women, there are only 4 seats, so 4P3. Now we multiply them together!

26. mathmate

For some reason, men have to fight for women! lol

27. dan815

remember its not 4 men and 4 women mathmate,

28. mathmath333

yes^ i was wondering that pic

29. mathmate

Four seat, but not all filled!

30. mathmate

First women gets to choose from 4, then 3, then 2.

31. mathmate

*woman

32. mathmate

One seat will be empty, so remove the chair and two men will sit next to each other.

33. dan815

yep

34. dan815

here is a more involved question

35. mathmate

Woah, not too difficult, I hope!

36. dan815

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

Is this a trick question? They seem like the ones we did.

38. dan815

no this one is more straightforward but theres some nice ways to do it efficiently

39. dan815

ill leave u the answers see if u get them both

40. mathmate

4!5P4 for the first one, and 3!4P4 for the second.

41. dan815

part a =1152 part b= 72

42. mathmate

So I got them both wrong! lol

43. mathmath333

for first question of round table is this possible|dw:1440447723500:dw|

44. dan815

oh oops ya, i thought the question said no 2 women and no 2 wmen can sit together

45. mathmath333

46. dan815

no i wanted it to be like that, no gender together xD

47. mathmath333

oh it is different

48. mathmate

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

yep thats right

50. dan815

51. mathmate

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

u r close

53. dan815

remember that now since its m and w on a round table mwmwmw = wmwmwm

54. dan815

so its not 1152/8 but 1152/2/8

55. mathmate

Oh, for the straight row, I have 1152=2*4!4! But for the round table, I still have 144=3!4!

56. dan815

well the way i connected the first answer to the 2nd is

57. dan815

u see how its 2*4!*4! theres no need for the 2 * anymore in the 2nd case since

58. dan815

w1m1w2m2 = m1w1m2w2 if u connect the edges its the same arrangement

59. dan815

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

w1m1w2m2 for example this is equal to m2w1m1w2

61. dan815

|dw:1440448675569:dw|

62. dan815

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

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

yep :)

65. dan815

do u get it too mathmath333

66. mathmate

Sorry, @mathmath333, we hijacked your post! :(

67. dan815

hehe

68. mathmath333

was offline ,lol

69. mathmate

Sorry mm333, we were discussing extensions to your problem! lol